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V 


ELEMENTARY 


G  E  O  M  E  T  RY 


I 


NRW    EDITION. 


INCLUDING    PLANE,    SOLID,     AND    SPHERICAL    GEOMETRY,     WITH 
PRACTICAL    EXERCISES. 


BY 


EDWARD    OLNEY, 

PROFESSOR     OF     MATHEMATICS    IN     TH^     UNIVERSITY     OF     MICHIGAN. 


SHELDON      AND      COMPANY, 

NEW    YORK    AND    CHICAGO. 

1883. 


OLNEY'S     NEW     SERIES 

EMBRACES  THE  FOLLOWING  BOOKS. 


FIRST  LESSONS  IN  ARITHMETIC, 
PRACTICAL  ARITHMETIC. 


Two-Book  Series. 


This  Series  has  more  examples,  and  at  less  price,  than  any  ever 
published.  _^«,,..,^ 


(^  strictly  High  School  Text-Book) 
Send  for  full  Circular  of  Olney's  Abithmeticd. 


FIRST  PRINCIPLES  OF  ALGEBRA. 

COMPLETE  ALGEBRA.    (Newly  electrotyped  in  large  type.) 
NEW  ELEMENTARY  GEOMETRY. 


Copyright,  1883,  by  Sheldon  &=  Co. 


Smith  &  McDougal,  Ei-kctbotvpekh. 
82  Buekmiii)  St.,  N.  Y. 


^i^^^l^     'l^  J  ^1^    'i^    '4^ 


THE  first  edition  of  Olney's  Special  or  Elementary 
Geometry  was  issued  nearly  twelve  years  ago.  It  con- 
tained many  new  features.  The  book  has  gone  into  use  in  every 
State  in  the  Union,  and  has  been  tested  by  practical  teachers  in 
all  grades  of  schools.  This  long  and  varied  test  has  been  watched 
with  care  by  the  author,  and  it  is  with  the  greatest  pleasure  that 
he  has  found  that  the  general  features  of  the  book  have  been 
well-nigh  universally  approved. 

To  make  the  book  still  more  acceptable  to  the  teachers  and 
schools  of  our  country,  and  to  keep  it  abreast  with  the  real 
advancement  in  science  and  methods  of  teaching,  as  well  as  to 
make  it  a  worthy  exponent  of  the  best  style  of  the  printer's  art, 
are  some  of  the  reasons  which  have  led.  to  the  preparation  of  this 
edition. 

1.  The  division  into  Chapters  and  Sections,  instead  of  Books, 
has  been  retained,  as  affording  better  means  of  classifying  the 
subject-matter,  and  also  as  conforming  to  the  usage  of  modern 
times  in  other  literary  and  scientific  treatises. , 

2.  Part  First  of  the  old  edition  has  been  omitted,  and  the 
definitions  and  illustrations  necessary  to  the  integrity  of  the 
subject  ha.ve  been  incorporated  with  the  body  of  the  work.  This 
has  been  done  solely  in  deference  to  the  general  sentiment  of  the 
teachers  of  our  country.  The  author  can  but  feel  that  this  senti- 
ment is  wrong.  That  the  best  way  to  present  the  subject  of 
Geometry  is  tb^present  some  of  its  leading  notions  and  practical 
facts  with  their  uses  in  drawing  and  in  common  life,  before 
attempting  to  reason  upon  them,  appears  to  him  quite  clear.  It 
is  in  accord  with  one  of  the  settled  maxims  of  teaching  which 


a  c  -i  -^  7 


IV  PREFACE. 

requires  "facts  before  reasoning,"  and  then  it  is  in  harmony 
with  the  historic  development  of  the  science,  and  with  the  order 
of  mental  development  in  the  individual.  Moreover,  since  this 
method  was  presented  to  the  American  public  in  this  treatise,  the 
author  has  received  books  on  exactly  the  same  plan,  which  are 
in  general  use  in  Germany,  and  also  "A  Syllabus  of  Plane 
Geometry,  prepared  by  the  Association  for  the  improvement  of 
Geometrical  teaching "  in  England,  in  which  this  principle 
is  recognized  by  recommending  quite  an  extended  course  in 
Geometrical  constructions  before  entering  upon  the  logical  treat- 
ment of  the  subject.  The  author  hopes  to  revise  his  Part  First, 
and  present  it  as  a  Httle  treatise  adapted  to  our  Grammar  or 
lower  schools ;  as  he  can  but  think  these  subjects  much  more 
interesting  and  usefuf  to  pupils  of  this  grade  than  much  of  the 
matter  usually  brought  before  them,  especially  the  more  advanced 
portions  of  arithmetic,  and  as  he  is  confident  that  they  are  the 
proper  preparation  for  the  intelligent  study  of  logical  geometry. 

3.  The  same  general  analysis  of  the  subject  is  adhered  to  as  in 
the  first  edition.  All  must  acknowledge  it  a  reproach  to  the 
oldest  and  most  perfect  of  the  sciences  that,  hitherto,  no  system- 
atic classification  of  its  subject-matter  has  been  reached.  That 
the  ordinary  arrangement  found  in  our  Geometries  is  not 
based  upon  a  scientific  analysis  of  the  subject,  and  a  systematic 
classification  of  topics  will  be  evident  to  any  one  who  attempts 
to  give  the  subject-title  of  almost  any  so-called  Book.  A 
glance  at  the  table  of  contents  of  this  volume  will  show  that 
the  analysis  of  the  subject-matter  is  simple  and  strictly  philo- 
sophical. There  are  two  lines  of  inquiry  in  geometry,  viz., 
concerning  position  (from  which  form  results)  and  magni- 
tude. The  concepts  of  Plane  Geometry  are  the  point,  straight 
line,  angle,  and  circle.  Kow,  the  measurement  of  magnitude  is 
either  direct  or  indirect.  The  direct  measurement  and  compar- 
ison of  magnitudes  is  a  simple  arithmetical  operation,  and  is 
presented,  as  regards  straight  lines,  in  Section  4.  The  direct 
measurement  of  other  magnitudes  is  effected  in  a  similar  manner, 
but  is  unimportant  from  a  scientific  point  of  view.  The  indirect 
measurement  of  magnitude,  as  when  we  find  the  third  side  of  a 
triangle  from  the  other  two  and  their  included  angle,  the  circum- 


PREFACE.  V 

ference  or  area  of  a  circle  from  the  radius,  etc.,  is  a  somewhat 
remote  application  of  more  elementary  principles.  There  is  then 
left,  as  the  natural  first  object  of  inquiry,  the  relative  position  of 
two  (and  hence  of  all)  straight  lines.  Here  we  have  philosophi- 
cally the  first  inquiry  of  logical  geometry.  This  inquiry  divides 
into  the  three  inquiries  concerning  perpendicular,  oblique  and 
parallel  lines.  In  a  similar  manner  the  topics  of  the  succeeding 
sections  unfold  themselves  from  the  principles  stated. 

4.  This  analysis  and  classification  of  the  subject-matter  re- 
quires that  a  somewhat  larger  number  of  propositions  be  demon- 
strated from  fundamental  principles,  that  did  the  old  method,  of 
proving  first  any  proposition  you  could,  and  then  any  other,  and 
so  on  ;  but  who  will  consider  this  a  defect  ?  On  the  other  hand, 
it  gives  almost  absolute  unity  of  method  of  demonstration  in 
the  propositions  of  any  one  section. 

5.  The  freedom  with  which  revolution  is  used  as  a  method  of 
demonstration,  will  be  observed  upon  a  cursory  reading.  Of 
course  it  is  assumed  that  the  old  repugnance  to  the  introduction 
of  the  notions  of  time  and  motion  into  geometry  is  outgrown. 
Indeed,  the  old  geometers  could  not  get  on  without  the  super- 
position of  magnitudes,  and  this  idea  involves  motion.  Now, 
revolution  is  but  a  systematic  method  of  effecting  superposition, 
which  is  well-nigh  the  only  geometrical  method  of  proving  the 
equality  of  magnitudes. 

6.  The  author  has  long  desired  to  introduce  the  idea  of  same- 
ness of  direction  in  treating  parallels ;  but  could  not  accept 
what  seemed  to  him  the  vague  methods  of  writers  who  have 
made  the  attempt.  If  we  cannot  define  the  notion  of  direction, 
we  certainly  should  have  some  method  of  estimating  and  measur- 
ing it  before  it  can  be  made  a  proper  subject  of  geometrical 
inquiry.  This  the  author  thinks  he  has  secured,  by  giving  the 
necessary  precision  to  certain  very  common  and  simple  notions. 

7.  As  to  the  introduction  of  the  infinitesimal  method  into 
mathematics  (and  if  introduced  at  all,  why  not  in  the  elements 
where  it  will  do  most  service  ?),  the  author  is  confident  that  no 
one  thing  would  do  more  to  simplify,  and  hence  to  advance, 
elementary  mathematical  study,  than  the  general  and  hearty 
acceptance  of  this  method.     No  writer  has  succeeded  in  getting 


VI  PREFACE. 

on  far,  even  in  pure  mathematics,  without  openly  or  covertly 
introducing  the  notion,  and  its  directness,  simplicity,  if  not 
absolute  necessity,  in  the  applied  mathematics  make  its  intro- 
duction into  the  elements  exceedingly  desirable.  Nevertheless, 
the  author  has  given  alternative  demonstrations,  either  in  the 
body  of  the  text  or  in  the  appendix,  so  that  those  who  prefer 
can  omit  the  demonstrations  involving  the  infinitesimal  concep- 
tion. 

8.  Thanks  to  the  spirit  of  the  times,  no  geometry  can  now 
receive  favor  which  does  not  give  opportunity  for  the  application 
of  principles  and  for  independent  investigation.  As  in  the 
former  edition,  so  in  this,  large  attention  has  been  given  to  this 
just  demand  of  the  times.  As  a  help  to  independent  thinking, 
after  the  student  has  been  fairly  introduced  to  the  methods,  and 
had  time  to  imbibe  somewhat  of  the  spirit  of  geometrical  reason- 
ing, the  references  to  the  antecedent  principles  on  which  state- 
ments in  the  demonstrations  are  based,  are  sometimes  omitted, 
and  their  place  supplied  by  interrogation  marks. 

9.  In  the  earlier  part  of  the  work,  the  demonstrations  are 
divided,  according  to  the  suggestion  originally  given  by  De 
Morgan,  into  short  paragraphs,  each  of  which  presents  but  a 
single  step.  So,  also,  in  this  part,  care  has  been  taken  to  make 
separate  paragraphs  of  the  statement  of  premises  and  the  conclu- 
sion, and  to  put  the  former  in  different  type  from  the  body  of 
the  demonstration.  But,  in  the  latter  part  of  the  work,  this 
somewhat  stiff  and  mechanical  arrangement  gives  place  to  the 
freer  and  more  elegant  forms  with  which  the  student  will  need  to 
be  familiar  in  his  subsequent  reading. 

10.  In  the  preparation  of  the  work  the  author  has  availed  him- 
self of  the  suggestions  of  a  large  number  of  the  best  practical 
teachers  in  all  parts  of  our  country.  His  chief  advisers  have 
been  Professor  Benjamin  F.  Clarke,  of  Brown  University,  K.  I., 
and  Professor  H.  N.  Chute,  of  the  Ann  Arbor  High  School, 
Mich.  To  Professor  Clarke  he  is  indebted  for  valuable  sugges- 
tions on  the  whole  of  Chapter  II.,  and  especially  on  triedrals. 
Indeed,  whatever  merit  there  may  be  in  the  general  method  of 
treatment  of  triedrals,  is  due  more  to  him  than  to  the  writer. 
His  ability  as  a  mathematician,  and  his  knowledge  of  what  is 


PREFACE,  yU 

practical  in  methods  of  presentation,  gained  by  long  experience 
in  teaching  the  subject,  appear  on  well-nigh  every  page  of  the 
latter  part  of  the  work.  Professor  Chute,  the  able  and  accom- 
plished teacher  of  geometry  in  the  Ann  Arbor  High  School,  has 
given  me  the  free  use  of  his  careful  and  scholarly  thought,  and 
long  and  successful  experience  as  a  teacher,  by  several  readings 
of  the  proofs,  and  by  the  use  of  the  advance  sheets  of  the  entire 
work  in  his  classes.  His  logical  acumen,  practical  skill,  and 
generous  contribution  of  whatever  he  has  found  most  valuable 
in  matter  or  method,  have  been  of  the  highest  service.  The 
same  general  acknowledgments  are  due  to  other  authors  as  were 
made  in  the  earlier  edition.  To  the  taste  and  skill  of  the  stereo- 
typers,  and  the  lavish  expenditure  of  patience  and  money  of  the 
PubHshers,  the  author  is  indebted  for  the  elegant  and  beautiful 
dress  in  which  the  book  appears. 

EDWAKD  OLNEY. 
University  op  Michigan, 

Ann  Arbor,  September  i,  1883. 

N.B. — Part  III.  of  the  old  edition  will  still  be  published  for  use  in  such 
schools  as  wish  to  push  the  study  of  geometry  still  further  than  it  is  carried 
in  the  ordinary  treatises,  and  especially  into  the  methods  of  what  is  called 
the  Modern  Geometry.  The  topics  embraced  in  that  part  are  Exercises  in 
Geometrical  Invention,  including  advanced  theorems  in  Special  or  Elemen- 
tary Geometry,  Problems  in  the  same,  and  Applications  of  Algebra  to 
Geometry  ;  and  also  an  Introduction  to  Modern  Geometry,  including  the 
elements  of  the  subjects  of  Loci,  Symmetry,  Maxima  and  Minima,  Isoperi- 
metry.  Transversals,  Harmonic  Proportion,  Pencils  and  Ratio,  Poles  and 
Polars,  Radical  Axes  and  Centres  of  Similitude  in  respect  to  Circles. 

The  author's  Trigonometry  can  also  be  had,  bound  separately  or  in  con- 
nection with  the  other  parts  of  the  Geometry,  the  same  as  formerly, 

E.  O. 


SUGGESTIONS  TO  TEACHERS. 

1.  Fix  firmly  in  mind  the  fundamental  definitions  of  the 
science,  in  exact  language,  and  illustrate  them  so  fully  that 
the  terms  cannot  be  used  in  the  hearing  of  the  pupil,  or  by 
him,  without  bringing  before  his  mind,  witJiout  conscious 
effort,  the  geometrical  conception. 

2.  By  numerous  and  varied  applications  of  the  fundamental 
principles  of  plane  geometry  to  the  most  famihar  and  homely 
things  in  common  life,  divest  the  pupil's  mind  of  the  impression 
that  he  is  studying  "  higher  mathematics  "  (as  he  is  not),  and 
beget  in  him  the  habit  of  seeing  the  applications  and  illustrations 
of  these  principles  everywhere  about  him. 

3.  By  means  of  much  experience  in  the  elements  of  geometri- 
cal drawing,  train  the  taste  to  enjoy,  the  eye  to  perceive,  and  the 
hand  to  execute,  geometrical  forms,  and  by  so  doing  fix  indelibly 
in  the  mind  the  "  working  facts  "  of  geometry. 

4.  Have  all  definitions,  theorems,  corollaries,  &c.,  memorized 
with  perfect  exactitude,  and  repeated  till  they  can  be  given  with- 
out effort.  Demonstrations  should  not  be  memorized  by  the 
pupil ;  and  considerable  latitude  may  be  allowed  in  the  use  of 
language,  provided  the  argument  is  brought  out  clearly.  But 
errors  in  grammar,  and  inelegancies  in  style,  should  be  carefully 
guarded  against.  One  of  the  chief  benefits  to  be  derived  from 
class-room  drill  in  mathematics  is  the  ability  to  think  clearly 
and  logically,  and  to  express  the  thought  in  concise,  perspicuous, 
and  elegant  language. 

5.  The  teacher  should  never  give  a  theorem  or  corollary  in 
proper  form,  but  by  some  such  half -questions  as  the  following, 
suggest  the  topic  : 

The  relation  between  the  hypotenuse  and  the  sides  of  a  right- 
angled  triangle  ? 

The  relative  position  of  two  circles  when  the  distance  between 
the  centres  is  less  than  the  sum  and  greater  than  the  difference 
of  the  radii  ? 


SUGGESTIONS   TO    TEACHERS.  ix 

The  sum  of  the  angles  of  a  triangle  ? 

The  relation  between  the  angles  and  the  sides  of  a  triangle  ? 
etc. 

In  this  manner  the  teacher  should  always  designate  the  propo- 
sition without  stating  it.  The  statement  is  one  of  the  most 
important  things  for  the  pupil  to  learn,  and  have  at  perfect  com- 
mand, and  hence  should  not  be  given  him  by  the  teacher. 

6.  The  construction  of  the  figure  is  a  necessary  part  of  the 
demonstration,  and  no  assistance  should  be  given  the  pupil,  nor 
aids  allowed. 

7.  All  figures  in  plane  geometry  should,  upon  first  going  over 
the  subject,  be  constructed  by  the  pupils  with  strict  accuracy,  on 
correct  geometrical  principles,  using  ruler  and  string ;  and  this 
should  be  persisted  in  until  it  can  be  done  with  ease.  In  reviews, 
free-hand  drawing  of  figures  may  be  allowed,  and  is  even  desir- 
able. 

8.  The  ordinary  notation  by  letters  should  be  used. 

9.  All  the  exercises  in  the  book  should  be  worked  with  care  in 
the  study,  and  in  the  class,  and  be  carefully  explained  by  the 
pupil ;  and  as  many  additional,  impromptu  exercises  as  may  be 
found  necessary  in  order  to  render  the  pupil  familiar  with  the 
practical  import  of  the  propositions. 

10.  Little,  if  any,  original  demonstration  of  theorems  not  in 
the  book  should  be  required  of  the  pupil  upon  first  going  over 
plane  geometry.  In  review,  more  or  less  of  such  work  may  be 
required. 

11.  Great  pains  should  be  taken  that  original  demonstrations 
be  given  in  good,  workmanlike  form.  For  this  purpose,  they 
should  be  written  out  with  care  by  the  pupil.  Indeed,  it  is  an 
excellent  occasional  exercise,  to  have  demonstrations  written 
out  in  full  in  class. 

12.  In  review,  much  attention  should  be  given  to  synopses  of 
demonstrations.  They  are  the  main  reliancie  for  fixing  in 
memory  the  line  of  argument  by  which  a  proposition  is  demon- 
strated. 


I 


INTRODUCTION. 
SECTION  I. 

GENERAL  DEFINITIONS  13-15 

SECTION  II. 
THE  GEOMETRICAL  CONCEPTS 16-^0 

SECTION   III. 
AXIOMS  AND  POSTULATES 30-33 

SECTION   IV. 
MEASUREMENT  OF  RIGHT  LINES 34-38 


CHAPTER  I. 

PLANE      GEOMETRY. 

SECTION  I. 

PERPENDICULAR  STRAIGHT  LINES 38-46 

Exercises 47-48 

SECTION   11. 

OBLIQUE  STRAIGHT  LINES 48-53 

Exercises 63-54 

SECTION  III. 

PARALLELS 55-66 

Exercises 67 

SECTION   IV. 

RELATIVE  POSITIONS  OF  STRAIGHT  LINES  AND  CIRCUMFERENCES     68-79 
Exercises 79 

SECTION   V. 

RELATIVE  POSITIONS  OF  CIRCUMFERENCES 80-86 

Exercises 87 


CONTENTS.  XI 

SECTION  VI. 

MEASUREMENT   OF  ANGLES 88-103 

Exercises 103-104 

SECTION   VII. 

ANGLES   OF    POLYGONS   AND   THE    RELATION    BETWEEN  THE  ANGLES 

AND  SIDES. 

Op  Triangles 104-109 

Of  QUADRn.ATERALS 110-118 

Of  Polygons  of  more  than  Four  Sides 118-121 

Of  Regular  Polygons 121-127 

Exercises 127-128 

Theorems  for  Original  Investigation 129-130 

SECTION   VIII. 

OF    EQUALITY. 

Op  Angles 130-133 

Op  Triangles 133-139 

Determination  op  Triangles 139-143 

Determination  op  Quadrilaterals 143-147 

Determination  of  Polygons 147-149 

Propositions  for  Original  Investigation 150-151 

SECTION    IX. 
OF  EQUIVALENCY  AND  AREAS 

Equivalency 151-155 

Area 155-163 

Exercises 163-164 

SECTION   X. 

OF  SIMILARITY 164-181 

Exercises 181-183 

SECTION   XI. 

APPLICATIONS   OF  THE   DOCTRINE   OF  SIMILARITY  TO  THE    DEVELOP- 
MENT  OF  GEOMETRICAL  PROPERTIES  OF  FIGURES. 

Op  the  Relations  op  the  Segments  on  two  Lines  in- 
tersecting EACH  OTHER  AND  INTERSECTED  BY  A  CIRCUM- 
FERENCE    184-186 

Op  THE  Bisector  op  an  Angle  of  a  Triangle.  . .     . .     186-188 

Areas  of  Similar  Figures  188-191 

Perimeters  and  the  Rectification  op  the  Circumper- 

ence 191-196 

Area  op  the  Circle 197-199 

Divison  in  Extreme  and  Mean  Ratio 199-201 

Exercises 201-204 


XU  CONTENTS. 

CHAPTER  II. 

SOLID      GEOMETRY. 

SECTION  I. 

OF  STRAIGHT  LINES  AND  PLANES.  p^gbs 

Plane  Determined 205-207 

Perpendicular  Lines  to  a  Plane. 208-213 

Oblique  Lines  to  a  Plane 213-216 

Parallel  Lines  to  a  Plane 216-218 

Parallel  Planes 218-223 

Exercises. 224 

SECTION  XL 
OF  SOLID  ANGLES. 

Of  Diedrals 225-231 

Of  Triedrals 231-244 

Of  Polyedrals 244 

Exercises 245-246 

SECTION  III. 

OF  PRISMS. AND  CYLINDERS 247-263 

Exercises 263-264 

SECTION  IV. 

OF  PYRAMIDS  AND  CONES 264-278 

Of  the  Regular  Polyedrons 279-282 

Exercises 282-283 

SECTION  V. 

OF  THE  SPHERE. 

Circles  of  the  Sphere 284-286 

Distances  on  the  Surface  op  a  Sphere 287-293 

Spherical  Angles 293-296 

Tangent  Planes 297-298 

Spherical  Triangles 298-309 

Polar  or  Supplemental  Triangles 309-312 

Quadrature  of  the  Surface  op  the  Sphere 313-316 

Lunes 317-323 

Volume  of  Sphere 324-327 

Spherical  Polygons  and  Pyramids 327-;J28 

Exercises 329-330 

APPENDIX 381-333 


/' 


^  E[eTnenCar 


V  ^:-  OR, 


-V 


yiEOMETI^ 


INTRODUCTION. 

PRELIMINARY    NOTIONS    AND    DEFINITIONS. 


^gCtlOH    I. 


GENERAL    DEFINITIONS.* 

1.  A  Proposition  is  a  statement  of  something  to  be  con- 
sidered or  done. 

Illustration. — Thus,  the  comraon  statement,  "Life  is  short,"  is  a 
proposition;  so,  also,  we  make,  or  state  a  proposition,  when  we  say, 
"Let  us  seek  earnestly  after  truth." — "The  product  of  the  divisor  and 
quotient,  plus  the  remainder,  equals  the  dividend,"  and  the  requirement, 
"  To  reduce  a  fraction  to  its  lowest  terms,"  are  examples  of  Arithmetical 
propositions. 

2.  Propositions  are  distinguished  as  Axioms,  Theorems,  Lem- 
mas, Corollaries,  Postulates,  and  Problems. 


*  The  terms  here  defined  are  such  as  are  used  in  the  science  in  conee- 
quence  of  its  logical  character,  hence  tliey  are  sometimes  called  logico- 
mathematical  terms.  The  science  of  the  Pure  Mathematics  may  be  con- 
sidered as  a  department  of  practical  logic. 


14  ELEMENTARY     GEOMETRY. 

3.  An  Axiom  is  a  proposition  which  states  a  principle  that 
is  so  elementary,  and  so  evidently  true  as  to  require  no  proof. 

Illustration. — Thus,  "  A  part  of  a  thing  is  less  than  the  whole  of 
it,"  "  Equimultiples  of  equals  are  equal,"  are  examples  of  axioms.  If  any 
one  does  not  admit  the  truth  of  axioms,  when  he  understands  the  terms 
used,  we  say  that  his  mind  is  not  sound,  and  that  we  cannot  reason  with 
him. 

4.  A  Theorem  is  a  proposition  which  states  a  real  or  sup- 
posed fact,  whose  truth  or  falsity  we  are  to  determine  by 
reasoning. 

Illustration. — "  If  the  same  quantity  be  added  to  both  numerator 
and  denominator  of  a  proper  fraction,  the  value  of  the  fraction  will  be 
increased,"  is  a  Theorem.  It  is  a  statement  the  truth  or  falsity  of  which 
we  are  to  determine  by  a  course  of  reasoning. 

6.  A  Demonstration  is  the  course  of  reasoning  by  means 
of  which  the  truth  or  falsity  of  a  theorem  is  made  to  appear. 
The  term  is  also  applied  to  a  logical  statement  of  the  reasons  for 
the  processes  of  a  rule. 

A  solution  tells  how  a  thing  is  done :  a  demonstration  tells  why  it  is 
so  done.     A  demonstration  is  often  called  proof. 

6.  A  Lemma  is  a  theorem  demonstrated  for  the  purpose 
of  using  it  in  the  demonstration  of  another  theorem. 

Illustration. — Thus,  in  order  to  demonstrate  the  rule  for  finding 
the  greatest  common  divisor  of  two  or  more  numbers,  it  may  be  best  first 
to  prove  that  "  A  divisor  of  two  numbers  is  a  divisor  of  their  sum,  and 
also  of  their  difference."  This  theorem,  when  proved  for  such  a  purpose, 
is  called  a  Lemma. 

The  term  Lemma  is  not  much  used,  and  is  not  very  important,  since 
most  theorems,  once  proved,  become  in  turn  auxiliary  to  the  proof  of 
others,  and  hence  might  be  called  lemmas. 

7.  A  Corollary  is  a  subordinate  theorem  which  is  sug- 
gested, or  the  truth  of  which  is  made  evident,  in  the  course  of 
the  demonstration  of  a  more  general  theorem,  or  which  is  a 
direct  inference  from  a  proposition,  or  a  definition. 

Illustration. — Thus,  by  the  discussion  of  the  ordinary  process  of 
performing  subtraction  in  Arithmetic,  the  following  Gorolhry  might  be 


GENERAL    DEFINITIONS.  15 

suggested :  "  Subtraction  may  also  be  performed  by  addition,  as  we  can 
readily  observe  what  number  must  be  added  to  the  subtrahend  to  pro- 
duce the  minuend." 

8.  A  Postulate  is  a  proposition  which  states  that  some- 
thing can  be  done,  and  which  is  so  evidently  true  as  to  require 
no  process  of  reasoning  to  show  that  it  is  possible  to  be  done. 
We  may  or  may  not  know  how  to  perform  the  operation. 

Illustration. — Quantities  of  the  same  kind  can  be  added  together. 

9.  A  Problem  is  a  proposition  to*do  some  specified  thing, 
and  is  stated  with  reference  to  developing  the  method  of  doing  it 

Illustration. — A  problem  is  often  stated  as  an  incomplete  sentence, 
as,  "To  reduce  fractions  to  forms  having  a  common  denominator." — This 
incomplete  statement  means  that  "  We  propose  to  show  how  to  reduce 
fractions  to  forms  having  a  common  denominator."  Again,  the  problem 
"  To  construct  a  square,"  means  that  "  We  propose  to  draw  a  figure 
which  is  called  a  square,  and  to  tell  how  it  is  done." 

10.  A  Rule  is  a  formal  statement  of  the  method  of  solving 
u  general  problem,  and  is  designed  for  practical  application  in 
solving  special  examples  of  the  same  class. 

11.  A  Solution  is  the  process  of  performing  a  problem  or 
an  example. 

A  solution  should  usually  be  accompanied  by  a  demonstration  of  the 
process. 

12.  A  Scholium  is  a  remark  made  at  the  close  of  a  dis- 
cussion, and  designed  to  call  attention  to  some  particular  feature 
or  features  of  it. 

Illustration.  —Thus,  after  having  discussed  the  subject  of  multipli- 
cation and  division  in  Arithmetic,  the  remark  that  "  Division  is  the  con- 
verse of  multiplication,"  is  a  scholium. 

13.  An  Hypothesis  is  a  supposition  made  in  the  state- 
ment of  a  proposition,  or  in  the  course  of  a  demonstration. 

The  Data  are  the  things  given  or  granted  in  a  proposition. 
The  Conclusion  is  the  thing  to  be  proved. 

The  data  of  a  proposition  and  the  hypotheses  are  the  same  thing. 


16 


ELEMENTARY     GEOMETRY, 


(i^^lrCtfiaM  IL 


THE    GEOMETRICAL    CONCEPTS.* 


Points  are  designated 


POINTS. 

14.   A  Point  is  a  place  without  size, 
by  letters. 

Illustration. — If  we  wish  to  designate  any  particular  point  (place) 
on  the  paper,  we  put  a  letter  by  it,  and 
sometimes  a  dot  in  it.     Thus,  in  Fig.  1,  the] 
ends  of  the  line,  which  are  points,  are  desig- 
nated as  "  point  A,"  "  point  D ;"  or,  simply, 
as  A  and  D.     The  points  marked  in  the  I 
line  are  designated  as  "  point  B,"  "  point  C," 
or  as  B  and  C.     F  and  E  are  two  points 
above  the  line. 


Fig.  I. 


LINES. 

16.   A  Line  is  the  path  of  a  point  in  motion. 

Lines  are  represented  upon  paper  by  marks  made  with  a  pen  or  pen- 
cil, the  point  of  the  pen  or  pencil  representing  the  moving  point. 

A  line  is  designated  by  naming  the  letters  written  at  its  ex- 
tremities, or  somewhere  upon  it. 

Illustration. — In  each  case  in  Fig.  2,  conceive  a  point  to  start  from 
A  and  move  along  the  path  indicated  by  the  mark  to  B.  The  path  thus 
traced  is  a  line.     Since  a  point  has  no  size,  a  line  hcts  no  breadth,  though 


*  A  concept  is  a  thing  thought  about ; — a  thought-object.  Thus,  in  Arith- 
metic, number  is  the  concept ;  in  Botany,  plants ;  in  Geometry,  as  will 
appear  in  this  section,  points,  lines,  surfaces,  and  solids.  These  may  also  be 
said  to  constitute  the  subject-matter  of  the  science. 


THE     GEOMETRICAL     CONCEPTS.  17 

the  marks  by  which  we  represent  lines  have  some  breadth.  The  first  and 
third  lines  in  the  figure  are  each  designated  as  "the  line  AB."  The  sec- 
ond line  is  considered  as  traced  by  a  point  starting  from  A  and  coming 


around  to  A  again,  so  that  B  and  A  coincide.  This  line  may  he  desig- 
nated as  the  line  AmnA,  or  AmnE.  In  the  fourth  case,  there  are  three 
lines  represented,  which  are  designated,  respectively,  as  AmB,  AnB,  and 
AcB ;  or,  the  last,  as  AB. 

16.  Lines  are  of  Two  Kinds,  Straight  and  Curved.  A 
straight  line  is  also  called  a  Eight  Line.  A  curved  line  is  often 
called  simply  a  Curve. 

17.  A  Straight  Line  is  a  line  traced  by  a  point  which 
moves  constantly  in  the  same  direction.     (See  46,  a.) 

The  word  "  line  "  used  alone  generally  signifies  a  straight  line. 

18.  A  Curved  Line  is  a  line  traced  by  a  point  which  con- 
stantly changes  its  direction  of  motion. 

Illustration. —Thus,  in  (1),  Fig.  2,  if  the  line  AB  is  conceived  as 
traced  by  a  point  moving  from  A  to  B,  it  is  evident  that  this  point  moves 
in  the  same  direction  throughout  its  course ;  hence  AB  is  a  straight  line. 
If  a  body,  as  a  stone,  is  let  fall,  it  moves  constantly  toward  the  centre  of 
the  earth  ;  hence  its  path  represents  a  straight  line.  If  a  weight  is  sus- 
pended by  a  string,  the  string  represents  a  straight  line. 

Considering  the  line  represented  by  AtB,  (3),  Pig.  2,  as  the  path  of  a 
point  moving  from  A  to  B,  we  see  that  the  direction  of  motion  is  con- 
stantly changing. 


18 


ELEMENTARY     GEOMETRY, 


Sometimes  a  path  like  that  rep- 
resented in  Fig.  3  is  called,  though 
improperly,  a  Broken  Line.  It  is  not 
a  line  at  all ;  that  is,  not  one  line  :  it 
is  a  combination  of  straight  lines. 


SURFACES. 

19.  A  Surface  is  the  path  of  a  line  in  motion. 

20.  Surfaces  are  of  Two  Kinds^  Plane  and  Curved. 

21.  A  Plane  Surface,  or  simply  a  Plane,  is  a  surface 
such  that  a  straight  line  passing  through  any  two  of  its  points 
lies  wholly  in  the  surface.  Such  a  surface  may  always  be  con- 
ceived as  the  path  of  a  straight  line  in  motion. 

Illustration. — Let  AB,  Fig.  4,  be  supposed  to  move  to  the  right,  so 
that  its  extremities  A  and  B  move  at  the 
same  rate  and  in  the  same  direction,  A 
tracing  the  line  AD,  and  B  the  line  BC.  The 
path  of  the  line,  the  figure  ABCD,  is  a  sur- 
face. This  page  is  a  surface,  and  may  be 
conceived  as  the  path  of  a  line  sliding  like  a 
ruler  from  top  to  bottom  of  it,  or  from  one 
side  to  the  other.     Such  a  path  will  have  '^'    ' 

length  and  breadth,  being  in  the  latter  respect  unlike  a  line,  which  has 
only  length. 

22.  A  Curved  Surface  is  a  surface  in  which,  if  various 
lin€8  are  drawn  through  any  point,  some  or  all  of  them  will  be 
curved. 

Illustration. — Suppose  a  fine  wire  bent  into  the  form  of  the  curve 
AmB,  Fig.  5,  and  its  ends  A  and  B  stuck  into  a  rod  XY.  Now,  taking 
the  rod  XY  in  the  fingers  and  rolling  it,  it  is  evident  that  the  path  of  the 
line  represented  by  the  wire  AwB  will  be  the  surface  of  a  ball  (sphere). 

Again,  suppose  the  rod  XY  placed  on  the  surface  of  this  paper  so 
that  the  wire  AmB  shall  stand  straight  up  from  the  paper,  just  as  it 


THE    GEOMETRICAL     CONCEPTS, 


19 


would  if  we  could  take  hold  of  the  curve  at  tn  and  raise  it  right  up, 
letting  XY  lie  as  it  does  in  the  figure.  Now  slide  the  rod  straight  up  or 
down  the  page,  making  both  ends  move  at  the  same  rate.    The  path  ol 


Fig.  6.  Fig.  7. 

A/?iB  will  be  like  the  surface  of  a  half-round  rod  (a  semi-cylinder).  Thus 
we  see  how  surfaces,  plane  and  curved,  may  be  conceiyed  as  the  paths  of 
lines  in  motion. 


Ex.  1.  If  the  curve  AnB,  Fig.  6,  be  conceived  as  revolved 
about  the  line  XY,  the  surface  of  what  object  will  its  path  be  like? 

Ex.  2.  If  the  figure  OMNP,  Fig.  7,  be  conceived  as  revolved 
about  OP,  what  kind  of  a  path  will  MN  trace  ?  What  kind  of 
paths  will  FN  and  OM  trace  ? 

Ans.  One  path  will  be  like  the  surface  of  a  joint  of  stove- 
pipe, L  e.,  a  cylindrical  surface  ;  and  one  will  be  like  a  flat  wheel, 
i.  e.,  a  circle. 

Ex.  3.  If  you  fasten  one  end  of  a  cord  at  a  point  in  the  ceil- 
ing and  hang  a  ball  on  the  other  end,  and  then  make  the  ball 
swing  around  in  a  circle,  what  kind  of  a  surface  will  the  string 
describe  ? 

Ex.  4.  If  on  the  surface  of  a  stove-pipe,  you  were  to  draw 
various  lines  through  the  same  point,  might  any  of  them  be 
straight  ?  Could  all  of  them  be  straight  ?  What  kind  of  a  sur- 
face is  this,  therefore? 

Ex.  5.  Can  you  draw  a  straight  line  on  the  surface  of  a  ball  ? 
On  the  surface  of  an  egg  ?     What  kind  of  surfaces  are  these  ? 


20  ELEMENTARY    GEOMETRY. 

Ex.  6.  When  the  carpenter  wishes  to  make  the  surface  of  a 
board  perfectly  flat,  he  takes  a  ruler  whose  edge  is  a  straight  line, 
and  lays  this  straight  edge  on  the  surface  in  all  directions, 
watching  closely  to  see  if  it  touches  at  all  points  in  all  positions. 
Which  of  our  definitions  is  he  illustrating  by  hig  practice  ^ 

Ex.  7.  How  can  you  conceive  a  straight  line  to  move  so  that 
it  shall  not  generate  a  surface  ? 


OF     TH  E    CI  RCLE. 

23.  A  Circle  is  a  plane  surface  bounded  by  a  curved  line 
all  points  in  which  are  equally  distant  from  a  point  within. 

24.  The  Circumference  of  a  circle  is  the  curved  line 
all  points  in  which  are  equally  distant  from  a  point  within. 

25.  The  Centre  of  a  circle  is  the  point  within,  which  is 
equally  distant  from  all  points  in  the  circumference. 

26.  An  Arc  is  a  part  of  a  circumference. 

27.  A  Radius  is  a  straight  line  drawn  from  the  centre  to 
any  point  in  the  circumference  of  a  circle. 

By  reason  of  (24)  all  radii  of  the  same  circle  are  equal. 

28.  A  Diameter  of  a  circle  is  a  straight  line  passing 
through  the  centre  and  limited  by  the  circumference. 

A  diameter  is  equal  to  the  sum  of  two  radii;  hence,  all  diam- 
eters of  the  same  circle  are  equal. 

Illustration. — A  circle  may  be  conceived  as  the  path  of  a  line,  like 
OB,  Fig.  8,  one  end  of  which,  0,  remains  at  the  same  point,  while  the 
other  end,  B,  moves  around  it  in  the  plane  of  the  paper.  OB  is  the  radius, 
and  the  path  described  by  the  point  B  is  the  circumference.  A  Bis  a  diam- 
eter. In  Fig.  9,  the  curved  line  ABCDA  is  the  circumference,  0  is  the 
centref&n^  the  surface  within  the  circumference  is  the  circle.     Any  part  of 


THE    GEOMETRICAL    CONCEPTS, 


31 


Fig.  8.  Fig.  ».  Fig.  10. 

a  circumference,  as  AB,  or  any  one  of  tlie  curved  lines  BB,  Fig.  8,  is  an 
arc.  So  also  AM  and  EF,  Fig.  10,  are  arcs.  EF  is  an  arc  drawn  from  0' 
as  a  centre,  with  the  radius  O'B. 


29.  A  Chord  is  a  straight  line  joining  any  two  points  in  a 
circumference,  as  BC  or  AD,  Fig.  9.  The  portion  of  the  circle 
included  between  the  chord  and  its  arc,  as  A/«D,  is  a  Segfiiient. 

30.  A  Tangent  to  a  circle  is  a  straight  line  which  touches 
the  circumference,  but  does  not  intersect  it,  how  far  soever  the 
line  be  produced. 

Two  circles  which  touch  each  other  in  but  one  point  are  said 
to  be  tangent  to  each  other.  A  straight-line  tangent  is  called  a 
Rectilinear  Tangent. 

31.  A  Secant  is  a  straight  line  which  intersects  the  circum- 
ference. 


ANGLES. 

32.  A  Plane  Angle,  or  simply  an  Angle,  is  the  opening 
between  two  lines  which  meet  each  other. 

The  point  in  which  the  lines  meet  is  called  the  Vertex,  and 
the  lines  are  called  the  Sides. 

An  angle  is  designated  by  pla.  lui,'  ;-  letter  at  its  vertex,  and 
one  by  each  of  its  sides.  In  reaciiiig.  we  name  the  letter  at  the 
vertex  when  there  is  but  one  vertex  at  the  point,  and  the  three 


5535  ELEMENTARY    GEOMETRY. 

letters  when  there  are  i^o  or  more  vertices  at  the  same  point. 
In  the  latter  case,  the  letter  at  the  vertex  is  put  between  the 
other  two. 

Illustration. — Id  com- 
mon language  an  angle  la 
called  a  comer.  The  open- 
ing between  the  two  lines 
AB  and  AC,  in  which  the 
ligure  1  stands,  is  called  the 
angle  A ;  or,  if  we  choose, 
we  may  call  it  the  angle 
BAC.  At  L  there  are  two 
vertices,  so  that  were  we  to 
say  the  angle  L,  one  would 
not  know  whether  we  meant 
the  angle  (corner)  in  which 
4  stands,  or  that  in  which 
6  stands.  To  avoid  this 
ambiguity,  we  say  the  angle 
HLR  for  the  former,  and 
RLT  for  the  latter.  The 
angle  ZAY  is  the  corner  in 

which  11   stands;   that  is,   ^ 

the  opening    between    the  '^' 

two  lines  AY  and  AZ.  In  designating  an  angle  by  three  letters,  it  is 
immaterial  which  letter  stands  first,  so  that  the  one  at  the  vertex  is  put 
between  the  other  two.  Thus,  PQS  and  SQP  are  both  designations  of 
the  angle  in  which  6  stands.  An  angle  is  also  frequently  designated  by 
putting  a  letter  or  figure  in  it  and  near  the  vertex. 

33.  The  Size  of  an  Angle  depends  upon  the  rapidity 
with  which  its  sides  separate,  and  not  upon  their  length. 

Illustration. —The  angles  BAC  and  MON,  Fig.  11,  are  equal,  since 
the  sides  separate  at  the  same  rate,  although  the  sides  of  the  latter  are 
more  prolonged  than  those  of  the  former.  The  sides  DF  and  DE  separate 
faster  than  AB  and  AC,  hence  the  angle  EDF  is  greater  than  the  angle 
BAC. 

34.  AdUacent  Angles  are  angles  so  situated  as  to  have  a 
common  vertex  and  one  common  side  lying  between  them. 


THE    GEOMETRICAL     CONCEPTS. 


Illustration. — In  Fig.  12,  angles  4  and 
5  are  adjdcenty  since  tliey  have  the  common 
vertex  L,  and  the  common  side  LR.  Angles 
9  and  10  are  also  adjacent. 

35.     Angles    are  distinguished    as 

Right    Angles     and  Oblique     Angles. 

Oblique    angles    are  either    Acute    or 
Obtuse. 


Fig.  12. 


36.  A  Right  Angle  is  an  angle  included  between  two 
straight  lines  which  meet  each  other  in  such  a  manner  as  to 
make  the  adjacent  angles  equal. 

37.  An  Acute  Angle  is  an  angle  which  is  less  than  a 
right  angle,  i.  e.,  one  whose  sides  sepamte  less  rapidly  than  those 
of  a  right  angle. 

38.  An  Obtuse  Angle  is  an  angle  which  is  greater  than  a 
right  angle,  i.  e.^  one  whose  sides  separate  more  rapidly  than  those 
of  a  right  angle. 

39.  A  Straight  Angle  is  an  angle  whose  sides  extend  in 
opposite  directions,  and  hence  form  one  and  the  same  straight  line. 

Illustrations. — In  comnjon  language,  a  right  angle  is  called  a 
square  corner,  and  an  acute  angle  a  sharp  comer. 


Fig.  13. 

Angles  BAD  and  BAC,  Fig.  13,  are  right  angles^  PST  is  an  acme  arigle^ 
and  HLR  is  an  obtuse  angle. 

If  HL  were  turned  to  the  left  until  it  fell  in  the  dotted  line,  the  angle 
HLR  would  increase,  and  when  HL  fell  in  the  dotted  line,  the  angle  would 
become  what  is  called  a  straight  angle. 


24  ELEMENTARY    GEOMETRY. 

40.  The  Sum  of  Two  Angles  is  the  angle  included 
between  their  non-coincident  sides,  when  the  two  angles  are  so 
placed  as  to  be  adjacent  angles,  and  their  sides  lie  in  the  same 
plane. 


Fig.  14. 

Illustration. — Let  0  and  M  be  any  two  angles.  Make  EPB  =  M, 
and  APE  =  0,  thus  placing  the  two  angles  0  and  M  so  that  they  become 
adjacent  angles  (34).     Then  is  APB  the  sum  of  0  and  M,  and  we  write, 

0  +  M  =  APB,        or        APE  +  EPB  =  APB. 

That  is,  the  sum  of  the  angles  0  and  M,  or  APE  and  EPB,  is  APB. 

41.  The  Difference  between  Two  Angles  is  the 

angle  included  by  their  non-coincident  sides,  when  the  angles 
are  so  placed  as  to  have  a  common  vertex  and  side,  the  second 
side  of  the  less  angle  lying  between  the  sides  of  the  greater. 


Fig.  15. 

Illustration.— To  find  the  difierence  between  the  two  angles  0  and 
S,  we  place  the  vertices  0  and  S  at  a  common  point,  as  at  P,  making 
APB  =  RST,  and  APC  =  DOE.  Then  is  CPB  the  difiference  between 
RST  and  DOE  ;  that  is, 

RST  -  DOE  =  CPB. 
So  also  APB  -  APC  =  CPB, 
and  APB  -  CPB  =  APC. 


THE    GEOMETRICAL    CONCEPTS.  25 

42.  Corollary  !.—(</)  The  sum  of  two  right  angles, 
(6)  Or,  the  sam  of  the  two  adjacent  angles  formed  by 
one  straight  line  meeting  another, 

(c)  Or,  the  sum  of  all  the  consecutive  angles  included 
by  several  lines  lying  on  the  same  side  of  a  given  line  and 
m^eeting  it  in  a  com^m^on  point,  is  a  straight  angle. 


Fig.  16. 

Thus,  ABP  +  PBC,   or  DEG  +  GEP,  or  HIL  +  LIM  +  MIN  +  NIK, 

is  a  straight  angle. 


43.  Corollary  2.— The  sum  of  the  four  angles  formed 
by  two  intersecting  lines,  or  the  sum.  of  all  the  consecutive 
angles  formed  by  any  numher  of  lines  meeting  in  a  com- 
mon point  is  two  straight  angles,  or  four  right  angles. 

Thus,  the  sum  of  the  four  angles  ADC,  CDB,  BDE,  and  EDA  is  four 
right  angles,  as  also  is  the  sum  of  AOB,  BOC,  COD,  DOE,  EOF,  FOG, 
and  GOA. 


26  ELEMENTARY    GEOMETRY, 

44.  A  Solid  is  a  limited  portion  of  space. 

Illustration. — Suppose  you  have  a  block  of  wood  like  that  repre. 
sented  in  Fig.  17.  Hold  it  still  in 
your  fingers  a  moment,  and  fix  your 
mind  upon  it.  Now  take  the  block 
away  and  think  of  the  space  (place) 
where  it  was.  This  space  is  an  ex- 
ample of  what  we  call  a  Solid  in 
Geometry.  In  fact,  the  solids  of 
Geometry  are  not  solids  at  all,  in 
the  common  sense  of  the  word 
solid ;  they  are  only  places  of  certain  shapes. 

Again,  hold  your  ball  still  a  moment  in  your  fingers,  then  let  it  drop, 
and  think  of  the  place  it  filled  when  you  had  it  in  your  fingers.  It  is 
this^Zace,  shaped  just  like  your  ball,  that  we  think  about  and  talk  about 
as  a  solid  in  Geometry. 


GENERATION    OF    LINES,    SURFACES, 
AND    ANGLES. 

45.  When  one  geometrical  concept  is  conceived  to  move  so 
that  its  path  is  some  other  concept,  the  former  is  said  to  generate 
the  latter,  and  the  latter  is  called  the  locus  of  the  former. 

The  Locus  of  a  Point  is  the  line  (either  straight  or 
curved)  generated  by  the  motion  of  the  point  according  to  some 
given  law. 

In  the  same  manner,  a  surface  is  conceived  as  the  locus  of  a 
line  moving  in  some  determinate  manner. 

46.  A  Line  is  generated  by  a  moving  point  (15-18).  Hence, 
the  locus  of  a  point  is  a  line. 


"TMB  g:eoMetrical  co^cjspts.  %^ 

(a)  The  came  straight  line  may  be  conceived  as  generated  by  a  point 
moving  in  either  of  two  opposite  directions,  or  part  of  it  may  be  con- 
ceived as  generated  by  a  point  moving  in  one  direction,  and  part  by  a 
point  moving  in  the  opposite  direction.  Thus,  FA,  Fig.  18,  may  be  con- 
ceived as  generated  by  a  point  moving  from  F  to  A,  or  from  A  to  F  ;  or 
the  part  OA  may  be  conceived  as  generated  by  a  point  moving  from  0  to 
A,  and  the  part  OF  by  a  point  moving  in  the  opposite  direction,  i.  <?., 
I  from  0  to  F. 

47.  A  Surface  is  generated  by  a  moving  line  (19-22). 
Hence,  the  locus  of  a  line  is  a  surface. 

48.  An  Angle  is  generated  by  the  revolution  of  a  straight 
line  about  one  of  its  extremities,  the  line  lying  all  the  time  in 
the  same  plane. 

Illustbation. — The  angle  BOA, 
Fig.  18,  may  be  considered  as  gen- 
erated by  the  revolution  of  the  line 
BO  from  the  position  AO  to  its  pres- 
ent position.  The  angle  COB  may 
be  considered  as  generated  by  the 
revolution  of  CO  from  the  position 
BO  to  its  present  position,  etc.  i^ 

49.  A  Right  Angle  is  generated  by  one-fourth  of  an  en- 
tire revolution,  an  Acute  Angle  by  less  than  one-fourth  of  an 
entire  revolution,  and  an  Obtuse  Angle  by  more  than  one- 
fourth.  A  Straight  Angle  is  generated  by  one-half  of  a 
revolution. 

50.  A  Solid  may  be  conceived  as  generated  by  the  motion 
of  a  plane,  and  hence  may  be  defined  as  the  path  of  a  plane  in 
motion. 

Illustration.— Thus  the  solid,  Fig.  17,  may  be  conceived  as  gener- 
ated by  the  movement  of  the  plane  ABCD  from  its  present  position  to  the 
position  GHFE. 

61.  A  Sphere  may  be  conceived  as  generated  by  the  revo- 
lution of  a  semi-circle  about  its  diameter.  (See  illustration  at 
the  bottom  of  page  18.) 


38  ELEMENTARY   GEOMETRY, 


QUERIES. 

■  1.  If  the  surface  OMNP,  Fig.  19,  is  conceived  as  reyolyed 
around  OP,  what  is  the  path  through  which  it  moves  ? 

Caution. — The  student  should  distinguish  between  the  surface  gener- 
ated by  the  line  MN,  and  the  solid  generated  by  the  surface  OMNP. 


Fig.  19.  Fig.  20. 

S5.  If  the  surface  represented  by  CAB,  Fig.  20,  is  conceived  as 
revolved  about  its  side  CA,  what  kind  of  a  solid  is  its  path  ? 

3.  As  you  fill  a  vessel  with  water,  what  is  the  solid  traced  by 
the  surface  of  the  water  ? 

Ans.  The  same  as  the  space  within  the  vessel. 

4.  If  a  circle  is  conceived  as  lying  horizontally,  and  then 
moved  directly  up,  what  will  be  the  solid  described,  i.  e.,  its  path  ? 
Do  not  confound  the  surface  described  with  the  solid.  What 
describes  the  surface  ?    What  the  solid  ? 


EXTENSION    AND    FORM. 

52.  Extension  means  a  stretching,  or  reaching  out. 
Hence,  a  Point  has  no  extension.    It  has  only  position  (place). 

A  Line  stretches  or  reaches  out,  but  only  in  length,  as  it  has 
no  width.  Hence,  a  line  is  said  to  have  One  Dimension,  viz., 
length. 

A  Surface  extends  not  only  in  length,  but  also  in  breadth ; 
and  hence  has  Two  Dimensions,  viz.,  length  and  breadth. 


THE    GEOMETRICAL    CONCEPTS,  29 

A  Solid  has  Three  Dimensions,  viz.,  length,  breadth,  and 
thickness. 

Illustration.  —Suppose  we  think  of  a  point  as  capable  of  stretching 
out  (extending)  in  one  direction.  It  would  become  a  line.  Now  sup- 
pose the  line  to  stretch  out  (extend)  in  another  direction — to  widen.  It 
would  become  a  surface.  Finally,  suppose  the  surface  capable  of  thick- 
ening, that  is,  extending  in  another  direction.     It  would  become  a  solid. 

63.  The  Limits  (extremities)  of  a  line  are  points. 
The  Limits  (boundaries)  of  a  surface  are  lines. 
The  Limits  (boundaries)  of  a  solid  are  surfaces. 

64.  Magnitude  (size)  is  the  result  of  extension.  lines, 
surfaces,  and  solids  are  the  geometrical  magnitudes.  A  point  is 
not  a  magnitude,  since  it  has  no  size.  The  magnitude  of  a  line 
is  its  length ;  of  a  surface,  its  area ;  of  a  solid,  its  volume. 

55.  Figure  or  Form  (shape)  is  the  result  of  position  of 
points.  The  form  of  a  line  (as  straight  or  curved)  depends  upon 
the  relative  position  of  the  points  in  the  line.  The  form  of  a 
surface  (as  plane  or  curved)  depends  upon  the  relative  position  of 
the  points  in  it.  The  form  of  a  solid  depends  upon  the  relative 
position  of  the  points  in  its  surface. 


QU  ER  I  ES 


1.  Suppose  a  line  to  begin  to  contract  in  length,  and  continue 
the  operation  till  it  can  contract  no  longer,  what  does  it  become  ? 
That  is,  what  is  the  minor  limit  of  a  line  ? 

2.  If  a  surface  contracts  in  one  dimension,  as  width,  till  it 
reaches  its  limit,  what  does  it  become  ? 

3.  If  a  solid  contracts  to  its  limit  in  one  dimension,  what  does 
it  pass  into  ?    If  in  two  dimensions  ?    If  in  three  dimensions  ? 

4.  What  kind  of  a  surface  is  that,  every  point  in  which  is 
equally  distant  from  a  given  point? 


30  ELEMENTARY     GEOMETRY. 

56.  Geometry  is  that  science  which  treats  of  magnitude 
and  form  as  the  result  of  extension  and  position. 

The  Geometrical  Concepts  are  points,  lines,  surfaces 
(including  plane  and  spherical  angles),  and  solids  (including 
solid  angles).* 

The  Object  of  the  science  is  the  measurement  and  compari- 
son of  these  concepts. 

Plane  Geometry  treats  of  figures  all  of  whose  parts  are  confined  to  one 
plane.  Solid  Geometry^  called  also  Geometry  of  Spaxx^  and  Geometry  of 
Three  IHmensions,  treats  of  figures  whose  parts  lie  in  different  planes. 
The  division  of  this  treatise  into  two  chapters  is  founded  upon  this  dis- 
tinction. 


0irct!0H  m 


AXIOMS     AND     POSTULATES, 

67.  There  are  very  many  axioms ;  but,  as  they  are  truths 
which  the  mind  grants  on  the  mere  statement,  it  is  not  needful 
to  enumerate  them  all.  We  give  a  few  of  the  more  important, 
with  some  illustrative  remarks. 

58.  All  demonstration  is  based  upon  definitions,  axioms,  or 
previously  demonstrated  propositions. 

*  A  plane  angle  may  be  conceived  as  a  portion  of  a  plane,  and  hence  as 
itself  a  surface,  and  thus  capable  of  increase  or  diminution  like  the  other 
magnitudes.  The  angle  thus  considered  becomes  a  sort  of  infinity  deter- 
mined relatively  by  the  rate  of  separation  of  the  lines.  It  is  thus  analogous 
to  an  infinite  series  the  law  of  which  is  determined  by  a  few  of  its  first  terms, 

^e  definitions  32,  33,  and  48,  with  their  illustrations. 


AXIOMS    AND    POSTULATES,  31 

69.  Axiom  I. — A  straight  line  is  the  slwrtest  line  be- 
tween two  points, 

Illustbation. — If  a  cord  is  stretched  across  the  table,  it  marks  a 
straight  line.  In  this  way  the  caipenter  marks  a  straight  line.  Having 
rubbed  a  cord,  called  a  chalk-line,  with  chalk,  he  stretches  it  tightly  from 
one  point  to  another  on  the  surface  upon  which  he  wishes  to  mark  the 
line,  and  then  raising  the  middle  of  the  cord,  lets  it  snap  upon  the  sur- 
face. So  the  gardener  makes  the  edges  of  his  paths  straight  by  sti-etching 
a  cord  along  them.  These  operations  depend  upon  the  principle  that 
when  the  line  between  the  points  is  the  shortest  possible,  it  is  straight. 


60.   Axiom  II. — Two  points  in  a  straight  line  deter- 
mine  its  position. 

Illustration. — If  the  farmer  wants  a  straight  fence  built,  he  sets  two 
stakes  to  mark  its  ends.  From  these  its  entire  course  becomes  known. 
This  is  the  principle  upon  which  aligning  (or  sighting)  depends.  Two 
points  in  the  required  line  being  given,  by  looking  from  one  in  the  direc- 
tion of  the  other,  we  look  along  a  straight  line,  and  are  thus  able  to  locate 
other  points  in  the 
line.  If  the  points  A 
and  B  are  marked,  by 
putting  the  eye  at  A 
and  looking  steadily 
towards  B,  we  can  tell  whether  D  and  E  are  in  the  same  straight  line  with 
A  and  B,  or  not.  So  we  can  observe  that  C  and  C"  are  not  in  the  line ; 
but  that  C  is.  This  process  of  discovering  other  points  in  a  line  with  two 
given  points  is  called  aligning,  or  sighting.  In  this  way  a  row  of  trees  is 
made  straight,  or  a  line  of  stakes  set.  It  is  the  principle  upon  which  the 
surveyor  runs  his  lines,  and  the  hunter  aims  his  gun.  In  the  latter  case, 
the  two  sights  are  the  given  points,  and  the  mark,  or  game,  is  a  third 
point,  which  the  marksman  wishes  to  have  in  the  same  straight  line  as 
the  sights. 

61.  Axiom  III. — Between  the  same  two  points  there 
is  one  straight  line,  and  only  one. 

Illustration.— Let  any  two  letters  on  this  page  represent  the  situa- 
tion of  two  points ;  we  readily  see  that  there  is  one,  and  only  one,  straight 
path  between  them.    A^ain,  let  a  comer  of  the  desk  represent  one  point 


32  ELEMENTARY    GEOMETRY. 

and  a  comer  of  the  ceiling  of  the  room  represent  another  point ;  we  per- 
ceive at  once  that,  if  a  point  is  conceived  to  pass  in  a  straight  line  from 
one  to  the  other,  it  will  always  trace  the  same  path.  In  short,  as  soon  as 
two  points  are  mentioned,  we  think  of  the  distance  between  them  as  a 
single  straight  line, — for  example,  the  centre  of  the  earth  and  the  centre 
of  the  sun. 

Once  more,  conceive  A  and  B,  Fig.  21,  to  be  two  points  in  the  path  of 
a  point  moving  from  A  in  the  direction  of  B.  Now  all  the  points  in  the 
same  direction  from  A  that  B  is,  are  in  this  path ;  and  any  point  out  of  this 
line,  as  C  or  C",  is  in  a  different  direction  from  A. 

In  this  manner  we  draw  a  straight  line  on  paper  by  laying  the  straight 
edge  of  a  ruler  on  two  points  through  which  we  wish  the  line  to  pass, 
and  passing  a  pen  or  pencil  along  this  edge. 


62.  To  Intersect  is  to  cross ;  and  a  crossing  is  called  an 
Intersection, 

63.  Corollary. — Two  straight  lines  can  intersect  in  hut 
one  point;  for,  if  they  had  two  points  common,  they  would 
coincide  and  not  intersect. 

Ex.  1.  A  railroad  is  to  be  run  from  the  town  A  to  town  B. 
Kit  is  made  straight^  through  what  points  will  it  pass  ?  Can  it 
pass  through  any  points  not  in  the  same  direction  from  A  that 
B  is? 

Ex.  2.  If  I  live  on  the  south  side  of  a  straight  railroad,  and 
my  friend  on  the  north  side,  but  five  miles  farther  east  and  two 
miles  farther  north,  and  the  road  from  my  house  to  his  is 
straight,  how  many  times  does  it  cross  the  railroad  ? 

Ex.  3.  Can  you  always  draw  a  straight  line  which  shall  cut  a 
curve  (whatever  curve  it  may  be)  in  two  points  at  least  ?    Try  it. 


64.  Axiom  IV. — The  whole  is  greater  than  any  of  its 
parts. 

65.  Axiom  V. — The  whole  is  equal  to  the  sum  of  all 
its  parts. 


AXIOMS    AND    POSTULATES,  33 

66.  Axiom  VI. — Things  which  are  equal  to  the  same 
thing  are  e^ual  to  each  other. 

67.  Axiom  VII. — //  equals  he  increased  or  dimin- 
ished equally,  the  results  will  he  equal. 

68.  Axiom  VIII. — //  unequals  he  increased  or  di" 
minished  equally,  the  greater  will  give  the  greater  result, 
i.  e.,  the  inequality  wUl  exist  in  the  same  sense. 


POSTULATES 


69.  Postulatel!  like  axioms,  are  very  numerous,  and  it  would 
be  useless  to  attempt  to  enumerate  them  all.  We  give  a  few 
simply  as  specimens. 

70.  Postulate  I.  —  A  line  can  he  produced  to  any 
length. 

71.  Postulate  II. — From^  any  point  a  straight  line 
can  he  drawn  to  any  other  point. 

72.  Postulate  III. — Geometrical  magnitudes  can  he 
added,  suhtracted,  multiplied,  or  divided. 

73.  Postulate  FV. — A  geometrical  figure  can  he  con- 
ceived as  moved  at  pleasure,  without  changing  its  size  or 
the  relation  of  its  parts  (shape). 

74.  Postulate  V.  — Any  nurriber  of  lines  can  he  drawn 
making  equal  angles  with  a  given  line, 

75.  Postulate  Yl.  —  With  any  point  as  a  centre,  a 
circumference  can  he  drawn  with  any  radius. 


34 


ELEMENTARY    GEOMETRY, 


jsiriCtlOM  lY 


MEASUREMENT    OF    RIGHT    LINES. 

76.  The  Measure  of  a  line  is  another  line  which  is  con- 
tained in  it  an  exact  number  of  times. 

77.  A  Common  Measure  of  two  or  more  lines  is  a  line 
which  measures  each  of  them. 

78.  Commensurable  Lines  are  lilies  which  have  a 
finite  common  measure. 

79.  The  Sum  of  Two  Lines  is  the  line  formed  by  uniting 
them  so  that  one  shall  be  the  prolongation  of  the  other. 

80.  The  Difference  between  Two  Lines  is  the  line 
which  remains  after  the  length  of  the  less  has  been  taken  from 
the  greater. 

81.  Problem. — To  measure  a  straight  line  with  the 
dividers  and  scale. 


Solution.— Let  AB,  Fig.  22,  be  the  line  to  be  measured.  Take  the 
dividers,  Fig.  2  (frontis- 
piece), and  placing  the 
sharp  point  A  firmly 
upon  the  end  A  of  the 
line  AB,  open  the  di- 
viders till  the  other  point 
B  (the  pencil  point)  just 
reaches  the  other  end  of 
the  line,  B.    Then  letting  Fig.  22. 

the  dividers  remain  open  just  this  amount,  place  the  point  A  on  the 
lower  end  of  the  left-hand  scale,  as  at  o,  Fig.  1  (frontispiece),  and  notice 
where  the  point  B  reaches.    In  this  case,  it  reaches  3  spaces  beyond  the 


MEASUREMENT    OF   RIGHT    LINES,  35 

figure  1.     Now,  as  this  scale  is  inches  and  tenths  of  inches,*  the  line  AB  is 
1.3  inches  long. 

Ex.  1.  What  is  the  length  of  CD  ?  Ans.  .15  of  a  foot. 

Ex.  2.  What  is  the  length  of  EF  ?        Ajis.  .75  of  an  inch. 

Ex.  3.  What  is  the  leng£h  of  GH  ?  Ans.  1\  inches. 

Ex.  4.  What  is  the  length  of  IK  ?  Ans.  .18  of  a  foot 

Ex.  5.  Draw  a  line  3  inches  long. 

Ex.  6.  Draw  a  line  2.15  inches  long. 

Ex.  7.  Draw  a  line  1.25  inches  long. 

Ex.  8.  Draw  a  line  .85  of  an  inch  long. 

[Note. — Suppose  a  fine  elastic  cord  were  attached  by  each  of  its  ends 
to  the  points  A  and  B  of  the  dividers;  when  they  were  opened  so  as  to 
reach  from  C  to  D,  Fig.  22,  the  cord  would  represent  the  line  CD.  Now 
applying  the  dividers  to  the  scale  is  the  same  as  laying  this  cord  on  the 
scale.  Without  the  cord,  we  can  imagine  the  distance  between  the  points 
of  the  dividers  to  be  a  line  of  the  same  length  as  CD.] 

Ex.  9.  Find  in  the  same  way  as  above  the  length  and  width 
of  this  page.  Also  the  distance  from  one  corner  (angle)  to  the 
opposite  one  (the  diagonal). 


82.  Problem. — To  find  the  suvi  of  two  lines. 

Solution.— To  find  the  sum  of  AB  and  CD,  If  first  draw  the  indefi- 
nite line  EX.  With  the  di- 
viders I  obtain  the  length 
of  AB,  by  placing  one  point 
on  A  and  extending  the 
other  to  B.  This  length  I 
now  lay  off  on  the  indefinite  ^'S*  23. 

line  EX,  by  putting  one  point  of  the  dividers  at  E  and  with  the  othei 
marking  the  point  F.     EF  is  thus  made  equal  to  AB.     In  the  same  man- 

*  The  next  scale  to  the  right  is  divided  into  lOths  and  lOOths  of  a  foot. 
Thus,  from  p  to  10  is  1  tenth  of  a  foot,  and  the  smaller  divisions  are  hun- 
dredths. 

f  These  elementary  solutions  are  sometimes  put  in  the  singular,  as  the 
more  simple  style. 


36  ELEMENTARY    GEOMETRY. 

ner,  taking  the  length  of  CD 
with  the  dividers,  I  lay  it 
off  from  F  on  the  line  FX. 
Thus  I  obtain 

EG  =  EF  +  FG  ^.     ,, 

=  AB  +  CD. 

Hence,  the  sum  of  AB  and  CD  is  EG. 

Ex.  1.    Find  the  sum  of  AB  and  EF,  Fig.  22. 
Ex.  2.    Find  the  sum  of  EF,  CD,  and  GH,  Fig.  22. 
Ex.  3.    Make  a  line  twice  as  long  as  CD,  Fig.  22.    Three 
times  as  long.  

83.  Problem.— To  find  the  difference  of  two  lines. 

Solution.— To  find  the  difference  of  AB  and  CD,  I  take  the  length  of 
the  less  line  AB  with  the  dividers; 
and  placing  one  point  of  the  dividers 
at  one  extremity  of  CD,  as  C,  make 
CE  =  AB.  Then  is  ED  the  differ- 
ence of  AB  and  CD,  since 

ED  =  CD  -  CE  =  CD 

Ex.  1.    Find  the  difference  of  IK  and  EF,  Fig.  22. 

Ex.  2.    Find  the  difference  of  GH  and  CD,  Fig.  22. 

Ex.  3.  Find  how  much  longer  IK,  Fig.  22,  is  than  the  sum 
of  EF,  Fig.  22,  and  CD,  Fig.  23. 

Ex.  4.  Find  the  difference  of  the  sum  of  AB  and  GH,  and 
the  sum  of  CD  and  EF,  Fig.  22. 


84.  Problem. — To  find  the  ratio  of  two  commensurable 
lines. 

Solution. — Let  AB  and  CD  (Fig.  25)  be  the  two  lines  whose  ratio  we 
seek. 

Apply  the  shorter  (AB)  to  the  longer  (CD)  as  many  times  as  the  latter 
will  contain  the  former.     If  A B  is  contained  an  integral  number  of  times 
(say  3,  or  m)  in  CD,  then  AB  is  a  common  measure  of  AB  and  CD,  and  we 
,         AB       1 
^^'^^CD^^S'^"^- 


MEASUREMENT    OF  RIGHT   LINES,  37 

But  if  the  shorter  is  not  contained  in  the  longer  an  integral  number  of 
times,  apply  it  as  many  times  as  it  is  contained,  and  note  the  remainder; 
thus,  AB  is  contained  in  CD  once,  with  a  remainder  oD. 

Now  apply  this  remainder,  «D,  to  AB  as  many  times  as  AB  will  con- 
tain it,  which,  in  this  case,  is  once  with  a  remainder  &B. 


Fig.  25. 

Again,  apply  this  remainder,  JB,  to  aD,  the  former  remainder.  In 
this  case,  it  is  contained  once  with  a  remainder  cD. 

Again,  apply  cD  to  JB.    It  is  contained  twice,  with  a  remainder  dE. 

^nally,  applying  dE  to  cD,  we  find  it  contained  3  times,  without  any 
remainder. 

Hence,  dE  is  the  common  measure  of  AB  and  CD. 

Calling  dE  the  unit  of  measure,  1,  we  have, 


dE  =  \] 

cD  =  ZdE  =  3; 

^i  =  2d)  =  6; 

M  =  })E  =  U  +  dE  =  1', 

aD  =  ac  +  cD  =  10 ; 

AB  =  A6  +  ftB  —  aD  ->r  ac  = 

17; 

CD  =  Co  +  aD  =  AB  +  aD  : 

=  27. 

Hence  the  lines  AB  and  CD  are  to  each  other  as  the  numbers  17  and 
27 ;  AB  is  ^  of  CD ;  or,  expressed  in  the  form  of  a  proportion, 

AB  _17* 
CD  ~  27  •     ' 

[Note. — This  process  will  be  seen  to  be  the  same  as  that  developed 
in  Arithmetic  and  Algebra  for  finding  the  greatest  or  highest  Common 
Measure  of  two  numbers.  See  Practical  Arithmetic,  p.  362,  and 
Complete  Algebra,  (137).] 

*  This  method  will  not  always  obtain  the  exact  ratio,  both  because  of 
the  imperfection  of  the  measurement,  and  because  some  lines  are  incommen- 
Burable  by  any  finite  unit,  as  will  appear  hereafter. 


38  ELEMENTARY    GEOMETRY. 


Fig.  26. 

Ex.  1.     Find,  as  above,  the  approximate  ratio  of  AB  to  CD. 

Ratio,  — . 
Ex.  2.     Find,  as  above,  the  approximate  ratio  of  CD  and  IK. 

Ratio,  -• 

D 

Ex.  3.     Find,  as  above,  the  approximate  ratio  of  EF  to  GH. 

Ratio,  jr« 
Ex.  4.     Find,  as  above,  the  approximate  ratio  of  EF  to  CD. 

Ratio,  ~ 


CONTINUOUS    VARIATION. 

85.  A  magnitude  is  said  to  vary  continuously  when  in 
passing  from  one  value  to  another  it  passes  through  all  interme- 
diate values. 

Illustbation. — Let  the  line  EF,  Fig.  26,  be  produced  by  placing  a 
pencil  at  F  and  tracing  the  line  to  the  right,  until  it  becomes  equal  to  IK. 
EF  has  thus  been  made  to  be  successively  of  all  intermediate  lengths  be- 
tween its  present  length  and  the  length  of  IK ;  i.  e.,  it  has  varied  continu- 
ously. 

In  like  manner,  an  angle  may  be  conceived  to  vary  continuously  from 
one  magnitude  to  another.  Thus,  in  Fig.  27,  the  angle  CPB  may  be  made 
greater  or  less  by  revolving  CP  about  P.  By  such  a  revolution  of  CP  the 
angle  CPB  may  be  conceived  to  vary,  or  grow,  continuoudy  till  it  becomes 
CPB. 


•*•  CHAPTER   i. 

•     •••••       •     •     ••        • 

PLANE      GEOMETRY. 


OF    PERPENDICULAR    STRAIGHT    LINES. 

86.  A  Perpendicular  to  a  given  line  is  a  line  which 
makes  a  right  angle  (36)  with  the  given  line. 

87.  An  Obliqvie  line  is  a  line  which  makes  an  oblique  angle 
with  a  given  line. 

PROPOSITION     I. 

88.  Theorem. — At  any  point  in  a  straight  line,  one 
perpendicular  can  be  erected  to  the  line,  and  only  one, 
which  shall  lie  on  the  same  side  of  the  line. 

Demonstration. 

Let  AB  represent  any  line,  and  P  be  any  point  therein. 

We  are  to  prove  that,  on  the  same  side 
of  AB,  there  can  be  one,  and  only  one, 
perpendicular  erected  to  AB  at  P. 

From  P  draw  any  oblique  line,  as  PC, 
forming  with  AB  the  two  angles  CPB  and 
CPA. 

Now,  while  the  extremity  P,  of  PC, 
remains  at  P,  conceive  the  line  PC  to  re-  pjg.  27. 

volve  so  as  to  increase  the  less  of  the  two 

angles,  as  CPB,  continuously.     Since  the  sum  of  CPB  and  CPA  remaillS 
constant,  CPA  will  diminish  continuously. 


40 


ELEMENTARY    GEOMETRY, 


Hence,  for  a  certain  position  of  CP,  as 
C'P,  these  angles  will  become  equal.  In 
this  position,  the  line  is  perpendicular  to 
AB  (36,  86).  Therefore,  there  can  be  one 
perpendicular,  C'P,  erected  to  AB  at  P. 

Again,  if  the  line  C'P  revolve  from  the 
position  in  which  the  angles  are  equal,  one 
angle  will  increase  and  the  other  diminish ; 
hence  there  is  only  one  position  of  the  line 
on  this  side  of  AB  in  which  the  adjacent  angles  are  equal. 

Therefore  there  can  be  only  one  perpendicular  erected  to  AB  at  P, 
which  shall  lie  on  the  same  side  of  AB.     Q.  e.  d. 

89.  Corollary  1. — On  the  other  side  of  the  line  a  second 
perpendicular,  and  only  one,  can  he  erected  from  the  same 
point  in  the  line, 

90.  Corollary  2. — //  one  straight  line  meets  another 
so  as  to  make  the  angle  on  one  side  of  it  a  right  angle,  the 
angle  on  the  other  side  is  also  a  right  angle. 


PROPOSITION    II. 

91.  Theorem. — //  two  straight  lines  intersect  so  as  to 
mahe  one  of  the  four  angles  formed  a  right  angle,  the 
other  three  are  right  angles,  and  the  lines  are  mutually 
perpendicular  to  each  other. 


Demonstration". 

Let  CD  intersect  AB,  making  CEB  a 
right  angle. 

We  are  to  prove  that  CEA,  A  ED,  and 
DEB  are  also  right  angles,  and  that  CD  is 
perpendicular  to  AB,  and  AB  to  CD. 

By  (90),  since  CEB  is  a  right  angle, 
CEA  is  also  a  right  angle. 

In  like  manner,  as  BE  meets  CD, 
making  CEB  aright  angle,  BED  is  a  right 
angle,  by  (90). 

Again,  since  DEB  is  right,  DE  meets 
AB,  making  one  angle  right;  hence  the 
Other,  AED,  is  right  also  (90).    Q.  e.  d. 


ii^^^^^^H 


Fig.  28. 


OF   PERPENDICULAR    STRAIGHT   LINES. 


41 


Finally,  since  CD  meets  AB,  making  AEC  a  right  angle,  CD  is  perpen- 
dicular to  AB  (86) ;  and,  since  AB  meets  CD,  making  AEC  a  right  angle, 
AB  is  perpendicular  to  CD.    Q.  e.  d. 


PROPOSITION     III. 

92.  Theorem.— TFT^en.  two  straight  lines  intersect  at 
right  angles,  if  the  portion  of  the  plane  of  the  lines  on  one 
side  of  either  line  he  conceived  to  revolve  on  that  line  as 
an  axis  until  it  coincides  with  the  portion  of  the  plane  on 
the  other  side,  the  parts  of  the  second  line  will  coincide,* 

Demonstration. 

Let  the  two  lines  AB  and  CD  intersect  at  right  angles  at  E ;  and  let 
the  portion  of  the  plane  of  the  lines  on  the  side  of  CD  on  which  B  lies 
be  conceived  to  revolve  on  the  line  CD  as  an  axis,  until  It  falls  In  the 
portion  of  the  plane  on  the  other  side  of  CD.f 

We  are  to  prove  that  EB  will  fall  in 
and  coincide  with  EA. 

The  point  E  being  in  CD,  does  not 
change  position  in  the  revolution;  and, 
as  EB  remains  perpendicular  to  CD,  it 
must  coincide  with  EA  after  the  revolu- 
tion, or  there  would  be  two  perpendicu- 
lars to  CD  on  the  same  side  and  from  the 
same  point,  E,  which  is  impossible  (88). 

Hence,  EB  coincides  with  EA.    Q.  e.  d. 

Fig.  29. 


•i^^^^^^H 


PROPOSITION     IV. 

93.  Theorem,— i^oTw.  any  point  without  a  straight 
line,  one  perpendicular  can  be  let  fall  upon  that  line,  and 
only  one. 

*  This  has  nothing  to  do  with  the  lengths  of  EB  and  EA  ;  indeed,  lines 
are  generally  supposed  indefinite  in  length,  unless  limited  by  the  data. 

f  This  revolution  may  be  illustrated  by  conceiving  the  paper  folded  in 
the  line  CD  until  EB  is  brought  into  EA. 


42 


ELEMENTARY    GEOGRAPHY, 


?' 

P 

\ 
\ 
\ 

\ 
— ) 

P' 

A     D 

/D"B     A'      D'       B' 

1  / 

/ 
/ 

jP 

Demoi^stration. 
Let  AB  be  any  line  and  P  any  point  without  the  line. 

We  are  to  prove  that  one  per- 
pendicular, and  only  one,  can  be 
let  fall  from  P  upon  AB. 

Let  A'B' be  an  auxiliary  line; 
and  at  any  point  in  it,  as  D',  let  a 
perpendicular  P'D'  be  erected  (88). 
Now  place  A'B',  bearing  P'D' 
with  it,  in  AB,  and  move  it  to  the 
right  or  left  till  P'D'  passes  through 
P,  and  when  in  this  position  let 
D  be  the  point  in  AB  in  which 
^''^-  3°-  D'  falls. 

Connect  P  and  D. 

Then,  since  angle  PDB  coincides  with  the  right  angle  P'D'B',  PDB  is  a 
right  angle,  and  PD  is  a  perpendicular  from  P  to  the  line  AB  (86)-  Q-  e.  d. 
We  are  now  to  prove  that  PD  is  the  only  perpendicular  from  P  to  the 
line  AB. 

Suppose  that  there  can  be  another,  and  let  it  be  PD". 
Produce  PD  to  P'",  and  take  DP"=DP,  and  draw  P"D". 
Now  let  the  portion  of  the  plane  above  AB  be  revolved  upon  AB  as 
an  axis  until  it  falls  in  the  plane  on  the  opposite  side  of  AB  from  its  first 
position.     Then  will  DP'  fall  in  DP'"  (92),  and  since  DP"  is  by  construc- 
tion equal  to  DP,  P  will  fall  in  P". 

Then,  since  PDB  is  a  right  angle  BDP"  is  also  a  right  angle,  and  PP" 
is  a  straight  line  (42,  a)- 

For  a  like  reason  PD"  P"  is  a  straight  line,  and  we  have  two  straight 
lines  from  P  to  P",  which  is  impossible. 

Hence  there  can  be  but  one  perpendicular,  as  PD,  from  P  upon  AB. 

Q.  E.  D. 


PROPOSITION    V. 

94.  Theorem.^From  a  point  laithout  a  straight  line, 
the  -perpendicular  is  the  shortest  distance  to  the  line. 

Demonstration. 
Let  AB  be  any  straight  line,  P  any  point  without  it,  PD  a  perpen- 
dicular, and  PC  any  oblique  line. 

We  are  to  prove  th^t  Pp  is  shorter  than  any  oblique  line,  as  PC. 


OF   PERPENDICULAR    STRAIGHT    LINES. 


43 


Ist.  Since  the  shortest  distance  from  P  to 
any  point  in  the  line  AB  is  a  straight  line 
(69),  we  are  to  examine  only  straight  lines. 

2d.  Produce  PD,  making  DP'  =  PD,  and 
draw  P'C. 

Now  let  the  portion  ofthe  plane  of  the  lines 
above  AB  be  revohed  upon  AB  as  an  axis  until 
it  coincides  with  the  portion  below  AB. 

Since  PP'  and  AB  intersect  at  right 
angles,  PD  will  fall  in  DP'  (92);  and,  since 
PD  =  DP',  P  will  fall  in  P',  and  PC  =  P'C, 
since  they  coincide  when  applied. 

Finally,  PP'  being  a  straight  line,  is  shorter  than  PCP'  which  is  a  broken 
line,  since  a  straight  line  is  the  shortest  distance  between  two  points  (59). 

Now  PD,  the  half  of  PP',  is  less  than  PC,  the  half  of  the  broken  line  PCP . 

Therefore,  the  perpendicular,  PD,  is  the  shortest  distance  from  R  to 
the  line  AB.     q.  e.  d. 

95.  The  Distance  between  two  points  is  the  straight  line 
which  joins  them,  and  the  Distance  from  a  point  to  a  line  is 
the  perpendicular  from  the  point  to  the  line. 


S^^IH^^E 


Fig.  31. 


PROPOSITION    VI. 

96.  Theorem. — //  a  perpendicular  is  erected  at  the 
middle  point  of  a  straight  line, 

1st.  Any  point  in  the  perpendicular  is  equally  distant 
from  the  extremities  of  the  line. 

2d.  Any  point  without  the  perpendicular  is  nearer  the 
extremity  ofthe  line  on  its  own  side  of  the  perpendicular. 

Demonstration^. 


Let  PD  be  a  perpendicular  to  AB  at 
its  middle  point,  D,  0  any  point  in  this 
perpendicular,  and  0'  any  point  without 
the  perpendicular. 

Draw  OA,  OB,  O'A,  and  O'B. 

We  are  to  prove,  1st,  that  OA  =  OB; 
and  2d,  that  O'B  <  O'A. 

Ist.  Revolve  ODB  on  PD  as  an  axis, 


Fig.  32. 


44  ELEMENTARY    GEOMETRY, 

till  B  falls  in  the  plane  on  the  opposite 
side  of  PD. 

Then,  since  PD  is  perpendicular  to  AB, 
DB  will  fall  in  DA  (92)-  And  since  DB 
=  DA  by  hypothesis,  B  will  fall  in  A,  and 
OB  will  coincide  with  OA  (61). 

Hence  OA  =  OB.    q.  e.  d. 

2d.  0'  being  on  the  opposite  side  of  PD 
from  A,  0' A  will  cut  PD  at  some  point,  as  0. 

Draw  CB.  ^''^-  ^^■ 

Now,  since  C  is  a  point  in  the  perpendicular,  CA  =  CB  by  the  for- 
mer part  of  the  demonstration. 

And,  since  O'B  is  a  straight  line  and  O'C  4-  CB  is  a  broken  line, 

OB  <  O'C  +  CB  (59). 
Whence,  substituting  CA  for  its  equal  CB,  we  have 
OB  <  O'C  +  CA, 
or        O'B  <  O'A.      Q.  E.  D. 

97.  Corollary. — Conversely,  The  locus  of  a  point  equi- 
distant from  the  extremities  of  a  given  line  is  a  perpen- 
dicular to  that  line  at  its  middle  point,  since  any  point  in 
such  perpendicular  is  equidistant  from  the  extremities  of  the  line, 
and  any  point  not  in  the  perpendicular  is  unequally  distant  from 
the  extremities. 


PROPOSITION    VII. 

98.  Theorem. — //  each  of  two  points  in  one  line  is 
equally  distant  from  the  extremities  of  another  line,  the 
former  line  is  perpendicular  to  the  latter  at  its  middle 
point. 

Demonstration. 

Every  point  equally  distant  from  the  extremities  of  a  straight  line  lies 
in  a  perpendicular  to  that  line  at  its  middle  point,  by  (97)-  But  two 
points  determine  the  position  of  a  straight  line.  Hence,  two  points,  each 
equally  distant  from  the  extremities  of  a  straight  line,  determine  the 
position  of  the  perpendicular  at  the  middle  point  of  the  line.    Q.  e.  d. 


OF   PERPENDICULAR    STRAIGHT   LINES. 


45 


PROPOSITION    VIII. 

99.  Problem. — To  erect  a  perpendicular  to  a  given 
line  at  a  given  point  in  the  line. 

SOL[JTION. 
Let  XY  be  the  given  line,  and  A  the  given  point. 

We  are  to  erect  a  perpendicu- 
lar to  XY,  at  A. 

From  A  lay  off  on  each  side 
equal  distances,  as  AC  =  AB. 

From  C  and  B  as  centres,  with 
a  radius  sufficiently  great  to  cause 
the  arcs  to  intersect  at  some  point 
without  XY,  describe  arcs  intersecting  at  0. 

Pass  a  line  through  0  and  A,  and  it  will  be  the  perpendicular  sought. 


Fig.  33. 


Demoxstkatiok  of  Solution. 

Since  OA  has  two  points,  0  and  A,  each  equally  distant  from  B  and 
C,  OA  is  a  perpendicular  to  BC  at  A,  its  middle  point  (98)* 

But  BC  coincides  with  XY ;  hence  OA  is  perpendicular  to  XY  at  A. 

100.  Definition. — To  Bisect  anything  is  to  divide  it  into 
two  equal  parts. 


PROPOSITION    IX. 
101.  Problem. — To  bisect  a  given  line. 


Solution. 
Let  AB  be  the  given  line. 

We  are  to  bisect  it,  that  is,  to  divide  it 
into  two  equal  parts. 

From  the  extremities  A  and  B  as  cen- 
tres, with  any  radius  sufficiently  great  to 
cause  the  arcs  to  intersect  without  the  line 
AB,  describe  arcs  intersecting  in  two  points, 
as  m  and  n. 


i^^^^^if^^^^: 


Fig.  34. 


46 


J^LEMBNTARt    GEOMETRY. 


Pass  a  line  through  m  and  n,  intersect- 
ing AB  at  0. 

Then  is  0  the  middle  point  of  AB,  and 
AO  =  OB. 

Demonstration  of  Solution. 

Since  the  line  mn  has  two  points,  m  and 
n,  eacli  equally  distant  from  A  and  B,  it  is 
perpendicular  to  AB  at  its  middle  point  (98) 


Fig.  34. 


PROPOSITION    X. 

102.   Problem. — From  a  point  without  a  given  line,  to 
let  fall  a  perpendicular  upon  the  line. 


Solution. 
Let  XY  be  the  given  line,  and  0  the  point  without  the  line. 

We  are  to  let  fall  a  perpendicular 
from  0  to  XY. 

From  0  as  a  centre,  with  a  radius 
sufficiently  great  to  cause  the  arcs  to  in- 
tersect, descrilje  an  arc  cutting  XY  in  two 
points,  as  B  and  C. 

From  B  and  C  as  centres,  with  a  ra- 
dius sufficiently  great  to  cause  the  arcs  to 
intersect  without  XY,  describe  arcs  in- 
tersecting at  some  point,  as  D.  Fig.  35. 

Pass  a  line  through  0  and  D,  meeting  XY  in  A.     Then  is  OA  the 
perpendicular  sought. 


Demonstration  of  Solution. 

OA  being  produced  through  D  has  two  points,  0  and  D,  each  equally 
distant  from  B  and  C,  and  hence  is  perpendicular  to  BC,  which  coincides 
with  XY.     Hence,  OA  passes  through  0  and  is  perpendicular  to  XY. 


OF  PERPENDICULAR    STRAIGHT   LlJSES, 


47 


QU  E  R  I  ES. 

103.  1.  In  the  solution  of  Proposition  IX,  is  it  necessary 
that  the  arcs  which  intersect  at  n  should  be  struck  with  the  same 
radius  as  those  which  intersect  at  ?w  ?  Is  it  necessary  that  the 
two  intersections  be  on  different  sides  of  AB  ? 

2.  In  the  solution  of  Proposition  X,  is  it  necessary  that  the 
intersection  D  should  fall  on  the  opposite  side  of  XY  from  0  ? 
Why  is  it  necessary  to  take  the  radius  with  which  these  arcs  are 
struck  greater  than  half  of  BC  ? 


EXERCISES. 

104.  1.  A  mason  wishes  to  build  a  wall  from  0  (Fig.  36),  in  the 
wall  AB,  '* straight  across"  (perpendicular)  to  the  wall  CD,  which 
is  8  feet  from  AB.  He  has  only  his  10-foot  pole,  which  is  subdi- 
vided into  feet  and  inches,  with  which  to  find  the  point  in  the 
opposite  wall  at  which  the  cross  wall  must  join.  How  shall  he 
find  it?    What  principle  is  involved? 


Fig.  36. 


Fig.  37. 


2.  Wishing  to  erect  a  line  perpendicular  to  AB  (Fig.  37)  at 
its  centre,  I  take  a  cord  or  chain  somewhat  longer  than  AB,  and 
fastening  its  ends  at  A  and  B,  take  hold  of  the  middle  of  the  cord 
or  chain  and  carry  it  as  far  from  AB  as  I  can,  first  on  one  side 
and  then  on  the  other,  sticking  pins  at  the  most  remote  points, 
as  at  P  and  P'.  These  points  determine  the  perpendicular  sought. 
What  is  the  principle  involved  ? 


48 


ELEMENTARY    GEOMETRY, 


3.  Bisect  a  line  by  making  marks  on  only  one  side  of  it. 

4.  With  a  measuring-tape  as  an  instrument,  how  would  you 
erect  on  the  shore  a  perpendicular  to  the  straight  bank  of  a  lake, 
at  a  given  point  in  the  bank  ? 


-♦-•-♦- 


0]&Ct!iaH  11 


OF      OBLIQUE      STRAIGHT      LINES. 

105.  The  Supplement  of  an  angle  is  the  angle  which 
remains  after  it  has  been  taken  from  a  straight  angle,  or  two 
right  angles. 

106.  Supplemental  Angles  are,  therefore,  two  angles 
whose  sum  is  a  straight  angle,  or  two  right  angles  (42,  h). 

107.  Vertical,  or  Opposite  Angles  are  the  non-adja- 
cent angles  formed  by  the  intersection  of  two  straight  lines. 


PROPOSITION    I. 

108.  Theorem.—  Vertical,  or  opposite  angles  are  equal. 

Demokstration. 
Let  AB  and  CE  intersect  at  D. 

We  are  to  prove  that  ADC  =  BDE,  and 
CDB    =ADE. 

ADC  +  CDB  =  a  straight  angle  (42,  ft); 
and  for  the  same  reason  CDB  +  BDE  =  a 
straight  angle. 

Hence,  ADC  +  CDB  =  CDB  +  BDE ;  and 
subtracting  CDB  from  each  member,  we  have 
ADC  =  BDE. 

In  like  manner,  CDB  +  BDE  —  BDE  +  ADE ;  whence,  CDB  =  ADE. 
q.  E.  D. 


Fig.  38. 


OF    OBLIQUE    STRAIGHT    LINES. 


40 


PROPOSITION    II. 

109.  Theorem. — //  two  supplemental  angles  are  so 
placed  as  to  he  adjacent  to  each  other,  the  two  sides  not 
common  fall  in  the  same  straight  line. 

Demonstration. 

Let  AOB  and  B'O'E'  be  two  supplemental 
angles,  and  let  B'O'E'  be  placed  so  as  to  be 
adjacent  to  AOB,  e.  e.^  as  BOE. 

We  are  to  prove  that  AE  is  a  straight  line. 

Before  considering  B'O'E'  as  placed  adja- 
cent to  AOB,  produce  AO  to  E,  forming  AE. 

By  (42,  ft),  AOB  +  BOE  =  a  straight 
angle,  i.  e.,  two  right  angles,  whence  BOE  is  the 
supplement  of  AOB. 

Now,  as  by  hypothesis  B'O'E'  is  the  supplement  of  AOB,  B'O'E' 
=  BOE. 

Place  B'O'E'  a^acent  to  AOB,  0'  in  0,  and  O'B'  in  OB.  Then  will 
O'E'  fall  m  OE. 

Therefore  the  two  sides  not  common,  i.  e.,  AO  and  O'E',  fall  in  the 
same  straight  line  AE-     Q.  e.  d. 


Fig.  39. 


PROPOSITION    III. 

110.  Theorem. — //  from,  a  point  without  a  line  a 
perpendicular  is  drawn  to  the  line,  and  oblique  lines  are 
drawn  from,  the  same  point,  meeting  the  line  at  equal 
distances  from,  the  foot  of  the  perpendicular, 

1st.  Tlie  oblique  lines  are  equal  to  each  other. 

2d.  The  angles  which  the  oblique  lines  form  ivith  the 
perpendicular  are  equal  to  each  other. 

3d.  The  angles  formed  by  the  oblique  lines  with  the  first 
line  are  equal  to  each  other. 

Demonstration. 
Let  AB  (Fig.  40)  be  any  line,  P  any  point  without  it,  PD  a  perpendicular, 
and  PC  and  PE  oblique  lines  meeting  AB  at  0  and  E,  so  that  CD  =  DE. 


50 


ELEMENTARY     GEOMETRY. 


We  are  to  prove,  1st,  that  PC  =  PE ; 
2d,  that  CPD  =  DPE ;  and  3d,  that 
PCD  =  PED. 

Revolve  PDE  on  PD  as  an  axis,  until 
E  falls  in  the  plane  on  the  other  side  of 
PD. 

Now,  since  AB  is  perpendicular  to 
PD,  DB  will  fall  in  DA  (92).  And  since 
DE  =  DC  by  hypothesis,  E  will  fall  in 
C.  Hence  the  two  figures  PDE  and  PDC 
coincide,  and  we  have,  1st,  PC  =  PE ;  2d,  CPD  =  DPE ;  and  3d,  PCD 
=  PED.     Q.  E.  D. 


Fig.  40 


Query.- 

(96)  y 


-How  would  the  equality  of  PC  and  PE  follo\y  from 


PROPOSITION    IV. 

111.  Theorem. — //  from  a  point  i^vithout  a  line  a 
perpendicular  is  draian  to  the  line,  and  from  the  same 
point  two  oblique  lines  are  drawn,  making  equal  angles 
with  the  perpendicular  and,  meeting  the  first  line, 

1st.  The  oblique  lines  are  equal  to  each  other. 

2d.  The  oblique  lines  cut  off  equal  distances  from  the 
foot  of  the  perpendicular. 

3d.  The  oblique  lines  make  equal  angles  with  the  first 

line.* 

Demonstratioit. 

Let  AB  be  a  straight  line,  P  any  point  without  it,  and  PD  a  perpen- 
dicular to  AB  ;  and  let  PE  and  PC  be  drawn,  making  CPD  =  EPD. 

We  are  to  prove,  1st,  that  PC  =  PE ; 
2d,  that  DE  =  DC;  and  3d,  that  PED 
=  PCD. 

Revolve  PDE  upon  PD  as  an  axis, 
until  E  falls  in  the  plane  on  the  opposite 
side  of  PD. 

Then,  since  EPD  =  CPD  by  hypoth- 
esis, PE  will  fall  in  PF,  and  the  point  E 
will  be  found  somewhere  in  PF.  c.     -, 

Fig.  41. 

*  This  proposition  is  the  converse  of  tlie  last.  The  significance  of  this 
statement  will  be  more  fully  developed  farther  on  (128). 


OF    OBLIQUE    STRAIGHT    LINES,  51 

Again,  DE  will  fall  in  DA  (92),  and  E  will  fall  somewhere  in  DA. 

Now  as  E  falls  at  the  same  time  in  DA  and  PF,  it  must  fall  at  their 
intersection  C,  and  the  figures  PDE  and  PDC  must  coincide ;  whence  we 
have, 

1st,  PC  =  PE ;  2d,  DE  =  DC ;  and  3d,  PCD  =  PED.     Q.  e.  d. 


PROPOSITION    V. 

112.  Theorem. — //  from  a  point  without  a  line  a 
perpendicular  is  drawn  to  the  line,  and  from  the  sam^e 
point  two  oblique  lines  are  drawn  making  equal  angles 
with  the  first  line, 

1st.  The  oblique  lines  cut  off  equal  distances  from  the 
foot  of  the  perpendicular, 

2d.   The  oblique  lines  are  equal  to  each  other. 

3d.  The  oblique  lines  make  equal  angles  with  the  per- 
pendicular. 

Demonstration. 

Let  P  be  any  point  without  the  line  AB,  and  PD  a  perpendicular  from 
P  upon  AB,  and  let  PE  and  PC  be  drawn  making  the  angle  DEP  =  angle 
DCP. 

We  are  to  prove,  1st,  that  DE  =  DC ; 
2d,  that  PE  =  PC ;  and  3d,  that  angle 
DPE  =  angle  DPC. 

Conceive  a  perpendicular  erected  at 
the  middle  point  of  CE,  and  let  it  intersect 
CP,  or  CP  produced,  in  some  point  as  X. 
Conceive  X  joined  with  E. 

By  (110,  3d.)  XED  =  XCD,  {i.  c,  PCD).  Fig.  42. 

But  by  hypothesis  PED  =  PCD.     Hence,  XE  falls  in  PE,  and  PD  is  the 
perpendicular  to  CE  at  its  middle  point. 

Therefore,  DE  =  DC;  and  by  (HO)  PE  =  PC,  and  DPE  =  DPC. 
Q.  E.  D. 


52  ELEMENTARY    GEOMETRY. 

PROPOSITION   VI. 

113.  Theorem. — If  from  a  point  without  a  line  a  per- 
pendicular is  let  fall  on  the  line,  and  from  the  same  point 
two  oblique  lines  are  drawn,  the  oblique  line  which  cuts  off 
the  greater  distance  from  the  foot  of  the  perpendicular  is 
the  greater. 

Demonstration. 

Let  AB  be  any  straight  line,  P  any  point  without  it,  and  PC  and  PF 
two  oblique  lines  of  which  PF  cuts  off  the  greater  distance  from  the  foot 
of  the  perpendicular  PD;  that  is,  DF  >  DC. 

We  are  to  prove  that  PF  >  PC. 

If  the  two  oblique  lines  do  not  lie  on 
the  same  side  of  the  perpendicular,  as  in 
the  case  of  PF  anfl  PE,  take  DC  =  DE, 
and  on  the  side  in  which  PF  lies, 
draw  PC.  Then  PC  will  be  equal  to  PE, 
by  (110,  1st).  Hence,  if  we  show  the 
proposition  true  when  both  oblique  lines 
lie  on  the  same  side  of  the  perpendicular, 
*  it  will  be  true  in  general.  ^'9*  *^' 

Produce  PD,  making  DP'  =  PD,  and  draw  P'F  and  P'C,  producing 
the  latter  until  it  meets  PF  in  H. 

Revolve  the  figure  FPD  upon  AB  as  an  axis,  until  it  falls  in  the  plane 
on  the  opposite  side  of  AB. 

Since  PP'  is  perpendicular  to  AB,  PD  will  fall  in  P'D;  and,  since 
PD  =  P'D,  P  will  fall  at  P'.     Then  P'C  =  PC  and  P'F  =  PF. 

Now  the  broken  line  PCP'  <  than  the  broken  line  PHP',  since  the 
straight  line  PC  <  the  broken  line  PHC. 

For  a  like  reason,  the  broken  line  PHP'  <  PFP',  since  HP'  <  HFP'. 

Hence  PCP'  <  PFP',  and  PC  (the  half  of  PCP')  <  PF  (the  half  of 
PFP').      Q.  E.  D. 

114.  Corollary. — From  a  given  point  without  a  line, 
there  can  he  two,  and  only  two,  equal  oblique  lines  draivn 
to  the  line,  and  these  will  lie  on  opposite  sides  of  the  per- 
pendicular  drawn  from  the  given  point  to  the  given  line. 


liH^^B! 


DF    OBLiqUE    STRAIGHT    LINES. 


PROPOSITION    VII. 

116.  Theorem.: — //  two  equal  oblique  lines  are  drawn 
from  the  same  point  in  a  perpendicular  to  a  given  line, 
they  cut  off  equal  distances  on  that  line  from,  the  foot 
of  the  perpendicular. 


Let  PD  be  perpendicular  to  AB,  and 
PE  =  PC. 

We  are  to  prove  that  DE  =  DC. 

If  DE  were  greater  than  DC,  PE  would 
be  greater  than  PC,  and  if  DE  were  less  than 
DC,  PE  would  be  less  than  PC  (113) ;  but 
both  of  these  conclusions  are  contrary  to 
the  hypothesis  PE  =  PC. 

Hence,  as  DE  can  neither  be  greater  nor 
equal  to  DC.     Q.  e.  d. 


Fig.  44. 
than  DC  it  must  l^e 


EXERCISES. 


116.  1.  Having  an  angle  given,  how  can  you  construct  its 
supplement  ?  Draw  on  the  blackboard  any  angle,  and  then  con- 
struct its  supplement.  What  is  the  supplement  of  a  right 
angle  ? 


Fig.  45. 


2.  The  several  angles  in  Fig.  45  are  such  parts  of  a  right 
angle  as  are  indicated  by  the  fractions  placed  in  them.  If  these 
anghfs  are  added  together  by  bringing  the  vertices  together  and 
causing  the  adjacent  sides  of  the  angles  to  coincide,  how  will  the 
two  sides  not  common  lie  ?    Why  ? 


54  ELEMENTARY    GEOMETRY. 

3.  If  two  times  A,  B  (Fig.  45),  two  times  D,  three  times  E, 
three  times  C,  three  times  G,  and  two  times  F,  are  added  in  order, 
how  will  AM  and  GN  lie  with  reference  to  each  other  ?     Why  ? 

Ans,  They  will  coincide. 

4.  If  you  place  the  vertices  of  any  two  equal  angles  together 
so  that  two  of  the  sides  shall  extend  in  opposite  directions  and 
form  one  and  the  same  straight  line,  the  other  two  sides  lying  on 
opposite  sides  thereof,  how  will  the  latter  sides  lie?  By  what 
principle  ? 

5.  If  two  lines  intersect,  show  that  tlie  line  which  hisects  one 
of  the  angles  will,  if  produced,  bisect  the  opposite  angle. 

6.  If  one  line  meet  another,  show  that  the  two  lines  bisecting 
these  supplemental  angles  are  perpendicular  to  each  other. 

7.  If  two  lines  intersect,  show  that  two  lines  bisecting  the 
two  pairs  of  opposite  angles  are  perpendicular  to  each  other. 


PARALLELS. 


I«» 


0irct!0H  m. 


OF     PARALLELS. 

117.  The  Direction  of  a  straight  line  is  defined  or  deter- 
mined by  the  plane  in  which  it  lies  and  the  angle  which  it  makes 
with  some  fixed  line,  this  angle  being  generated  (48)  from  the 
fixed  line  around  in  the  same  direction,*  in  the  same  argument. 


118.   The  assumed  fixed  line  is  called  the  Direction  Line, 

and  the  angle  which  the  line  makes  with  the  direction  hne  is 
called  the  Direction  Angle. 

Illustration. — Thus  the  directions 
of  the  several  lines  AB,  CD,  and  EF  may 
be  defined  by  referring  them  to  some  as- 
sumed fixed  line,  as  XY. 

The  direction  of  AB  is  defined  by  say- 
ing that  its  direction  angle  is  YOA,  or  its 
equal  XOB,  this  angle  l)cing  conceived  as 
generated /r^'7/i  the  direction  line,  as  indi- 
cated by  the  arrows. 

So  also  the  direction  of  CD  is  defined 
by  the  angle  YMC,  or  its  equal  XMD;  and 
the  direction  of  EF  is  in  like  manner  de- 
fined by  YNE,  or  XNF. 


Fig.  46. 


119.  Witli  reference  to  its  generation,  the  same 
line  may  be  conceived  as  having  either  of  two  opposite  directions, 
or  various  parts  of  it  may  be  conceived  as  having  opposite  direc- 
tions. 

Illustration.— Thus,  the  line  AB  (Fig.  47)  may  be  considered  as 
generated  by  a  point  moving  from  A  to  B,  whence  its  direction  would 

*  Revolution  around  a  fixed  point  is  often  designated  as  from  left  to 
right,  or  from  right  to  left.  To  comprehend  these  terms,  one  may  conceive 
himself  in  the  centre  of  motion,  and  facing  the  moving  point.  Thus  all  the 
motions  represented  by  arrows  in  Fig.  46  will  be  seen  to  he  from  right  to  left. 


56 


ELEMENTARY    GEOMETRY. 


be  from  A  towards  B ;  or,  it  may  be  considered 
as  generated  by  a  point  moving  from  B  to  A, 
whence  its  direction  would  be  from  B  towards  A. 
In  like  manner,  part  of  the  line,  as  PB,  may 
be  considered  as  having  the  direction  from  P 
towards  B,  while  the  other  part  is  conceived  as 
having  the  opposite  direction,  i.  e.,  from  P  towards  A. 


Fig.  47. 


120.  Lines  have  the  Same  Direction  when  they  lie  in 
the  same  plane  and  make  equal  direction  angles  with  the  same 
line. 

Any  line  may  be  assumed  at  pleasure  as  the  direction  line, 
provided  that  in  comparing  the  directions  of  different  Hnes  they 
all  be  referred  to  the  same  direction  line. 

121.  Parallel  Lines  are  lines  which  have  the  same  or 
opposite  directions. 

.122.   A  Transversal  is  a  line  cutting  a  system  of  lines. 


123.   When  two  lines  are  cut  by  a  transversal,  the  angles 
formed  are  named  as  follows : 


Exterior  Angles  are  those  without 
the  two  lines,  as  1,  2,  7,  and  8. 

Interior  Angles  are  those  within 
the  two  lines,  as  3,  4,  5,  and  6. 

Alternate  Exterior  Angles  are 

those  without  the  two  lines  and  on  differ- 
ent sides  of  the  transversal,  but  not  adja- 
cent, as  2  and  7,  1  and  8. 


Fig.  48. 


Alternate  Interior  Angles  are  those  within  the  two 
lines  and  on  different  sides  of  the  transversal,  but  not  adjacent, 
as  3  and  6,  4  and  5. 

Corresponding  Angles  are  one  without  and  one  within 
the  two  lines,  and  on  the  same  side  of  the  transversal,  but  not 
adjacent,  as  2  and  6,  4  and  8, 1  and  5,  3  and  7. 


PARALLELS.  67 

PROPOSITION    I. 

124.  Theorem. — Through  a  given  point  one  line  can  be 
drawn  parallel  to  a  given  line,  and  but  one. 

Demonstration. 
Let  AB  be  the  given  line,  and  P  the  given  point. 

We  arc  to  prove  that  one  line  can 
be  drawn  through  P  parallel  to  AB, 
and  but  one. 

Through  P  draw  XY  as  the  direc- 
tion line,  intersecting  AB  in  E. 

Also  through  P  pass  a  line  C'D', 
making  XPD'  greater  than  XEB. 

Then  revolving  CD'  about  P  as  a 
centre,  XPD'  may  be  made  to  dimin- 
ish continuously,  and  in  some  posi-  Fig.  49. 
tion,  as  CD,  XPD  will  equal  XEB.     In  this  position,  CD  is  parallel  to  AB 
(120,  121). 

Hence  there  can  be  one  line  drawn  through  P  parallel  to  AB.     q.  e.  d. 

Again,  there  can  be  but  one;  since,  if  CD  be  revolved  in  either  direc- 
tion about  P,  the  angle  XPD  will  become  unequal  to  XEB,  and  hence  the 
line  CD  will  not  be  parallel  to  AB.     Q.  e.  d. 


PROPOSITION    II. 

125.   Theorem. — //  a  transversal  cuts  two  parallels, 

1st.  Any  two  corresponding  angles  are  equal. 

2d.  Any  two  alternate  interior,  or  any  two  alternate 
exterior  angles  are  equal. 

3d.  The  sum  of  any  two  interior  angles  on  the  same 
side  of  the  transversal,  or  the  sum  of  any  two  exterior 
angles  on  the  same  side,  is  two  right  angles,  or  a  straight 
angle. 

Demonstration. 

Let  AB  and  CD  (Fig.  50)  beany  two  parallels,  and  EF  any  transversal. 


58 


ELEMENT  A  RT    GEO  METR  F. 


We  are  to  prove,  Ist.  Of  the  cor- 
responding angles,  b  =  d^  a  =  e, 
e  =  g^  and  /  =  h. 

2d.  Of  the  alternate  interior  an- 
gles, b  =^f,  and  c  =  g;  of  the  alter- 
nate exterior  angles,  d  =  h,  and 
a  =  e. 

3d.  Of  the  interior  angles  on  the 
same  side  of  the  transversal, 

&  +  c  =  2  right  angles, 
and     g  -\-  f  =  2  right  angles ; 

of  the  exterior  angles  on  the  same  side, 
a  +  d  =  2  right  angles,        and 


Fig.  50. 

e  +  h  =  2  right  angle 


Let  EF  be  taken  as  the  direction  line,  the  direction  angles  being  esti- 
mated from  right  to  left  (120,  121,  and  foot-note,  p.  55).     Then, 

1st.  Of  the  corresponding  angles,  b  =  d,  these  being  the  direction 
angles,  and  AB  and  CD  being  parallel. 

a  =  c,  since  they  are  supplements  of  the  equal  angles,  b  and  d ;  and 
e  =  g,  for  the  same  reason. 

Also,  f  =  h^  since  they  are  opposite  angles  to  the  equal  angles,  b 
and  d. 

2d.  Of  the  alternate  interior  angles,  ft  =/,  since  f  —  d  (108)  \  c  —  g^ 
since  they  are  supplements  of  b  and  d. 

Of  the  alternate  exterior  angles,  d  =  h^  since  h  =  b  (108);  and 
«  =  a,  since  they  are  supplements  of  b  and  d. 

3d.   Of  the  interior  angles  on  the  same  side,  « 

b  +  c  =  2  right  angles  (or  a  straight  angle), 
since       d  -\-  c  =  2  right  angles  (or  a  straight  angle),  (42,  *),  and  b  =  d; 

and  1 

since 


g  +  f  z=z  2  right  angles, 

g  -{.  b  =  a  straight  angle,  and  b  =f. 


Of  the  exterior  angles  on  the  same  side, 

a  +  d  =  2  right  angles, 
since  a  +  b  =  b,  straight  angle,  and  b  =  d; 

also 
since 


e  -\-  h  =  2  right  an^ 

g  -\-  h  =  &  straight  angle,  and  e  =  g.      Q.  e.  d. 


PARALLELS,  69 


PROPOSITION    III. 

126.  Theorem. — Conversely  to  Proposition  II,  When 
two  lines  are  cut  by  a  transversal,  the  two  lines  are  par- 
allel, 

|-      1st.  //  any  two  corresponding  angles  are  equal, 

2d.  //  a?nj  two  alternate  interior,  or  any  two  alternate 
exterior  angles  are  equal. 

3d.  //'  the  sum  of  any  two  interior  angles  on  the  same 
side,  or  the  sum  of  any  two  exterior  angles  on  the  same 
side  is  two  right  angles. 

Demonstration. 

Let  AB  and  CD  be  two  lines  cut  by  the  transversal  EF,  making  any 
pair  of  corresponding  angles  equal,  as  6  =  f/,  a  =  r,  y  =  e,  h  =  /'; 
or  any  two  alternate  interior  angles,  or  any  two  alternate  exterior 
angles  equal,  as  ft  =  /,  f/  —  c,  (i  =  e,  or  h  =  d;  or  the  sum  of  any 
two  interior  angles  on  the  same  side,  or  of  any  two  exterior  angles  on 
the  same  side,  equal  to  2  right  angles,  3S  h  +  c,  y  +  f,  a  ^  d,  h  +  e, 
equal  to  2  right  angles. 

Wc  are  to  prove  that  AB  and  CD 
are  parallel. 

Let  EF  be  the  direction  line,  and  b 
and  d  the  direction  angles.  If,  then, 
these  are  granted  or  proved  equal,  the 
lines  are  parallel  (121)- 

Now,  1st.  Of  the  corresponding  an- 1 
gles,  if  h  =  d,  AB  and  CD  are  parallel 
by   definition;    but,   if  a  =  c,  h  =  d, 
since  &  and  d  arc  supplements  of  a  and 

c;  or,  if  ^  =  e,  b  =  d,  since  b  and  d      "      "  Fig.  51. 

are  supplements  of  g  and  <?;  or,  if  ^  =/,  h  =  d^  since  h  =  h  and  d  =/ 
(108).    Hence,  in  every  case,  h  =  d,  and  AB  and  CD  are  parallel. 

2d.  Of  alternate  interior  angles,  if  ft  =  /,  b  =  d,  since  /  =  d;  or,  if 
g  =  c,  b  —  d,  since  b  and  d  are  supplements  of  g  and  c.  Hence,  in  either 
case,  b  =  d.  and  AB  and  CD  are  parallel. 


60  ELEMENTARY    GEOMETRY. 

Of  alternate  exterior  angles,  if 
h  =  d,I}  =  dy  since  b  =  h  (108) ;  o^i 
if  a  =  e,  J  =  df,  since  ft  and  d  are 
supplements  of  a  and  e.  Heuc^,  in 
either  case,  h  =  d  and,  AB  and  CD 
are  parallel. 

3d.  Of  interior  angles  on  the 
same  side,  if  &  +  «  =  3  right  angles, 
h  =  d,  since  d  +  c  =  2  right  angles 
(42) ;  or,  if  ^  +  /  =  2  right  angles, 
l>  =  d,  since  Pig,  5,^ 

g  +  f  +i  +  c  =  A  right  angles  ; 

hence,  ft  +  c  =  2  right  angles,  and  as  <f  +  c  =  2  right   angles,  d  =  h. 
Hence,  in  either  case,  ft  =  <f ,  and  AB  and  CD  are  parallel. 

Of  exterior  angles  on  the  same  side,  if  a  +  <Z  =  2  right  angles,  h  =  d, 
since  a  -{-  b  =  2  right  angles ;  or,  if  h  -\-  e  =  2  right  angles,  b  =  d,  since 
h  +  e-\-a-{-d  =  4:  right  angles;  hence,  a  ^  d  =  2  right  angles,  and,  as 
a  +  ft  =  2  right  angles,  d  =  b.  Hence,  in  either  case,  ft  =  (?,  and  AB  and 
CD  are  parallel,     q.  E.  D. 

127.  Corollary. — Two  lines  which  are  perpendicular 
to  a  third  are  parallel  to  each  other. 

For,  in  such  a  case,  all  the  eight  angles  formed  are  equal ;  hence,  any 
of  the  conditions  of  the  proposition  are  met. 

128.  Scholium. — The  last  two  propositions  are  the  converse 
of  each  other;  i.e.,  the  hypotheses  and  conclusions  are  exchanged. 
Thus,  in  Prop.  II,  the  hypothesis  is  that  the  two  lines  are  parallel, 
and  the  conclusion  is  certain  relations  between  the  angles ;  while 
in  Prop.  Ill  the  hypotheses  are  certain  relations  among  the  angles, 
and  the  conclusion  is  that  the  lines  are  parallel. 

The  learner  may  think  that,  if  a  proposition  is  true,  its  converse  is 
necessarily  true ;  and  hence,  that  when  a  proposition  has  been  proved,  its 
converse  may  be  assumed  as  also  proved.  Now  this  is  by  no  means  the 
case.  Although  in  a  great  number  of  mathematical  propositions,  it  hap- 
pens that  the  proposition  and  its  converse  are  both  true,  we  never  assume 
one  from  having  proved  the  other  ;  and  we  shall  occasionally  find  a  prop- 
osition whose  converse  is  not  true. 


PARALLELS. 


ei 


PROPOSITION    IV, 

129.  Theorem.  -  When  two  straight  lines  are  cut  hy  a 
transversal,  if  the  sum,  of  the  two  interior  angles  on  either 
side  is  less  than  two  right  angles,  tiie  two  lines  will  meet 
on  this  side  of  the  transversal,  if  sufficiently  extended. 

Demoi^stration. 

Let  AB  and  CD  be  two  lines  cut  by  the  transversal  XY,  making 
BEP  +  EPD  <  2  right  angles. 


We  are  to  prove  that  AB  and  CD  will  meet  on  the  side  of  XY  on  which 
these  angles  lie. 

Through  P  draw  FG  parallel  to  AB. 

Take  EH  =  EP  and  draw  PH,  and  also  ET  perpendicular  to  PH  By 
(115),  TH  =  TP.  whence  EHT  =  EPT  (110) 

But  EHT  =  GPH    125)      Hence  GPH  =  ^GPE. 

Again,  take  HI  —  PH  and  draw  PI,  and  it  may  be  shown  in  the  same 
manner  that  GPI  -=  ^GPH  ==  ^GPE. 

In  this  manner  we  may  continue  to  draw  oblique  lines  through  P  cut- 
ing  AB  further  and  further  from  E,  and  may  thus  diminish  at  pleasure 
the  angle  included  by  the  oblique  line  and  PG.  Hence  this  angle  may  be 
made  less  than  GPD,  the  difference,  between  DPE  and  the  supplement 
of  PEB,  when  the  oblique  line  will  fall  between  PD  and  PG.  Call  this 
line  PR.  Now  as  PR  and  PE  cut  AB,  and  PD  lies  between  them,  it  must 
cat  AB  between  E  and  R.    q.  e.  d. 


6S5  ELEMENTARY     GEOMETRY, 

130.  Corollary  1. — If  a  transversal  cuts  one  of  two 
parallels,  it  cuts  the  other  also. 

131.  Corollary  ^^i. — JVon-jMirallel  straight  lines  meety 
if  sufficiently  produced. 

132.  Corollary  3. — Two  straight  lines  in  the  same 
plane  which  do  not  meet,  however  far  they  are  extended, 
are  parallel. 

For,  let  AB  and  CD  be  two  such  lines,  and  P  any  point  in  CD.  Now 
all  lines  through  P  which  are  not4)arallel  to  AB  meet  AB  (131).  Hence 
as  there  can  be  one  parallel  to  AB  through  P  (124),  it  is  the  line  which 
does  not  meet  AS. 


PROPOSITION    V. 

133.    Theorem. — A  line  which  is  perpendicular  to  one 
of  two  parallels  is  perpendicular  to  the  other  also. 


Demonstration. 
Let  AB  and  CD  be  two  parallels,  and  let  EF  be  perpendicular  to  AB. 

We  are  to  prove  that  EF  is  also 
perpendicular  to  CD. 

Since  EF  is  a  transversal  cut- 
ting AB  and  CD,  angle  EOB  =  angle 
EMD  (125,  1). 

Now  EOB  is  a  right  angle  by 
hypothesis  (86), 

Hence  EMD  is  a  right  angle, 
and   EF   is    perpendicular    to    CD. 

Q.  E.  D. 

Fig.  53. 

134.   Corollary. — The    shortest  distance    between    two 
parallels  is  the  -perpendicular  which  joins  them. 

For,  CM  being  a  perpendicular  from  O  to  CD,  is  shorter  than  any 
other  line  from  0  to  CD  (94). 


PARALLELS.  63 

135.    The  Distance  between  two  parallels  is  the  perpen- 
dicular which  joins  them. 


• 


PROPOSITION    VI. 

136.  Theorem. — Two  parallels  are  everywhere  equally 
distant  from  each  other,  and  hence  never  meet. 

Demonstration. 

Let  E  and  F  be  any  two  points  in  the  line  CD,  and  EG  and  FH  per- 
pendiculars measuring  the  distances  between  the  parallels  CD  and  AB  at 
these  points. 

We  are  to  prove  EG  =  FH. 

Let  P  t>e  the  middle  point  between 
E  and  F,  and  PO  a  perpendicular  at 
this  point. 

Revolve  the  portion  of  the  figure 
on  the  right  of  PO,  upon  PO  as  an  axis,  f 'q-  54. 

until  it  falls  upon  the  plane  of  the  paper  at  the  left. 

Then,  since  FPO  and  EPO  are  right  angles,  PD  will  fall  in  PC ;  and, 
as  PF  =  PE,  F  will  fall  on  E.  As  F  and  E  are  right  angles,  FH  will 
take  the  direction  EG,  and  H  will  lie  in  EG  or  EG  produced.  Also, 
as  POH  and  POG  are  right  angles,  OB  will  fall  in  OA,  and  H,  falling  at  the 
same  time  in  EG  and  OA,  is  at  their  intersection  G. 

Hence,  FH  coincides  with  and  is  equ.il  to  EG.     Q.  e.  d. 

Hence,  also,  CD  cannot  meet  AB,  since  the  distance  from  any  point 
in  CD  to  AB  is  EG.     Q.  e.  d. 


PROPOSITION    VII. 

137.  Theorem. —Conversely  to  Profwsition  VI.,  //  two 
points  in  one  straight  line  are  equally  distant  from  a 
second  straight  line,  and  on  the  same  side  of  it,  the  lines 
are  parallel  to  each  other. 

Demonstration. 

Let  AB  and  CD  (Fig.  55)  be  two  lines  having  the  points  P  and  S  in  CD 
equally  distant  from  AB,  and  on  the  same  side  of  it. 

We  are  to  prove  that  CD  and  AB  are  parallel. 


64 


ELEMENTARY    GEOMETRY. 


From  P  and  S  draw  PE  and  SF  perpendicular  to  AB.  Then  is  PE  = 
SF,  by  hypothesis. 

Through  0,  the  middle  point  of  PE, 
draw  GH  parallel  to  AB. 

Since  PE  is  parallel  to  SF,  GH  cuts 
SF  in  some  point  as  I  (130)- 

By  (136),  OE  =  IF ;  and  since  SF  = 
PE  and  OE  is  |PE,  IF  is  ^SF,  that  is, 
IF  =  IS.  *''S-  55. 

Now,  as  PE  and  SF  are  perpendicu-ar  to  GH  (133),  if  we  revolye  the 
figure  OAEFBI  on  GH,  E  will  fall  in  P,  and  F  in  S  (92),  and  AB  will  have 
two  points  in  common  with  CD,  and  hence  will  coincide  with  it. 

Hence,  DPO  =  BEO,  and  as  the  latter  is  a  right  angle  by  construc- 
tion, AB  and  CD  are  perpendicular  to  PE,  and  hence  parallel  (127). 

Q.  E.  D. 


PROPOSITION    VIII. 

138.   Theorem.— t/i  pair  of  parallel  transversals  inter- 
cept equal  portions  of  two  parallels. 


Demonstratioi^. 

Let  ST  and  RL  be  two  parallel  transversals,  cutting  the  two  parallels 
AB  and  CD. 

We  are  to  prove  that  GE  =  HF. 
From  E  and  F  let  fall  the  per- 
pendiculars EM  and  FK.     Then 

EM  =  FK  (136). 

Now  apply  the  figure  GEM  to 
HFK,  placing  EM  in  its  equal  FK. 
Since  M  and  K  are  right  angles,  MG 
will  fall  in  KH. 

With  the  figures  in  this  position, 
FH  and  EG  are  lines  drawn  from  the 
same  point  in  the  perpendicular  to 
ST  and  making  equal  angles  with  it 
(125),  and  are  hence  equal  (112).     Q-  E.  D, 


Fig.  56. 


PARALLELS. 


«6 


PROPOSITION    IX. 

139.   Theorem. — Two  straight  lines  which  are  parallel 
to  a  third  are  parallel  to  each  other. 


Demonstration. 
Let  AB  and  CD  be  each  parallel  to  EF 

We  are  to  prove  that  AB  and  CD  are 
parallel  to  each  other. 

Draw  HI  perpendicular  to  EF;  then  will 
it  be  perpendicular  to  CD  (133)- 

For  a  like  reason,  HI  is  perpendicular 
to  AB. 

Hence,  CD  and  AB  are  both  perpendic- 
ular to  HI,  and  consequently  parallel  (127)-    Q.  k.  d 


<» 


Fig.  57. 


PROPOSITION    X. 

140.   Theorem. — //  to  each  of  two  parallels  perpendic- 
ulars are  drawn,  then  are  the  perpendiculars  parallel. 


Demonstration. 
Let  A  and  B  be  parallel  lines,  P  be  perpendicular  to  A,  and  Q  to  B. 

We  are  to  prove  that  P  and  Q  are  par- 
allel to  each  other. 

Q,  which  is  perpendicular  to  B,  one  of 
the  two  parallels,  is  perpendicular  to  A, 
the  other  parallel,  also  (133). 

Hence,  P  and  Q  are  both  perpendicu- 
lar to  A,  and  hence  are  parallel  (127)-    Q.  e.  d.| 

Fig.  58. 


66 


ELEMENT  A  R  Y     GEOMETR  T. 


PROPOSITION   XI. 

141.  Theorem. — //  to  each  of  two  non-parallel  lines 
a  perpendiGwlar  is  drawn,  the  perpendiculars  are  non- 
parallel. 

Demonstration. 

Let  A  and  B  be  non-parallel,  and  P  a  perpendicular  to  A,  and  Q  to  B. 

We  are  to  prove  that  P  and  Q  are 
non-parallel. 

If  P  and  Q  were  parallel,  then,  by  the 
preceding  proposition,  A  and  B  would  be 
parallel,  which  is  contrary  to  the  hy- 
pothesis. Hence,  P  and  Q  are  non-par- 
allel.    <i.  B.  D.  

Fig.  59. 


PROPOSITION    XII. 

142.    Problem. — Through  a  given  point  to  draw  a 
parallel  to  a  giveri  line. 

Solution. 
Let  AB  be  the  given  line,  and  P  the  given  point. 


Fig.  60. 

"We  are  to  draw  through  P  a  parallel  to  AB. 

Let  fall  PF,  a  perpendicular  from  P  to  AB  (102). 

At  P  erect  CD,  a  perpendicular  to  PF  (99). 

Then  is  CD  parallel  to  AB  (127).     [Pupil  give  proof.] 


PARALLELS. 


67 


EXE  RCISES. 

143.  1.  How  can  a  farmer  tell  whether  the  opposite  sides 
of  his  farm  are  parallel  ? 

2.  If  we  wish  to  cross  over  from  one  of  two  parallel  roads  to 
the  other,  is  it  of  any  use  to  travel  farther  in  the  hope  that  the 
distance  across  will  be  less  ?     Why  ? 

3.  If  a  straight  line  intersects  two  parallel  lines,  how  many 
angles  are  formed  ?  How  many  angles  of  the  same  size  ?  May 
they  all  be  of  the  same  size  ?  When  ?  Wlien  will  they  not  be 
all  of  the  same  size  ? 

4.  Are  the  two  opposite  walls  of  a  building  which  are  carried 
up  by  the  plumb  line  exactly  parallel  ?     Why  ? 

6.  A  hetad  (Fig.  61)  is  an  instni-l 
ment  much  used  by  carpenters,  and[ 
consists  of  a  main  limb  AB,  in 
which  a  tongue  CD  is  placed,  so  as  1 
to  open  and  shut  like  the  blade  of 
a  knife.  This  tongue  turns  on  the 
pivot  0,  which  is  a  screw,  and  can 
be  tightened  so  as  to  hold  the 
tongue  firmly  at  any  angle  with  the 
limb.  The  tongue  can  also  be  ad- 
justed so  as  to  allow  a  greater  ori 
less  portion  to  extend  on  a  given 
side,  as  CB,  of  the  limb.  Now,' 
suppose  the  tongue  fixed  in  posi- 
tion, as  represented  in  the  figure,  and  the  side  m  of  the  limb  to  be  placed 
against  the  straight  edge  of  a  board,  and  slid  up  and  down,  while  lines 
are  drawn  along  the  side  n  of  the  tongue.  What  will  be  the  relative 
position  of  these  lines  ?  Upon  what  proposition  does  their  relative  posi- 
tion depend  ?  How  can  the  carpenter  adjust  the  bevel  to  a  right  angle 
upon  the  principle  in  Prop.  I,  Sec.  I  ?  At  what  angle  is  the  bevel  set, 
when,  drawing  two  lines  from  the  same  point  in  the  edge  of  the  board, 
one  with  one  edge  m  of  the  bevel  against  the  edge  of  the  board,  and  the 
other  with  the  other  edge  m',  these  lines  are  at  right  angles  to  each  other? 


Fig.  61. 


68 


ELEMENTARY    GEOMETRY, 


^^0]^cti0M  ty 


OF  THE    RELATIVE    POSITIONS   OF  STRAIGHT    LINES  AND 
CIRCUMFERENCES. 


PROPOSITION    I. 

144.   Theorem. — Any  diameter  divides  a  circle^  and 
also  its  circumference,  into  two  equal  parts. 


Demon^stkation. 
Let  AB  be  the  diameter  of  the  circle  kin^n. 

We  are  to  prove  that  arc  A»iB  =  arc 
AwB,  and  that  segment  AwiB  =  segment  AwB. 

Revolve  A^B  upon  AB  as  an  axis,  until 
it  falls  in  the  plane  on  the  opposite  side  of  AB. 

Then,  since  every  point  in  AwB  is  at  the 
same  distance  from  the  centre  C  as  every 
point  of  A?nB  (24),  the  arc  AwB  falls  in  AwB, 
and  both  arcs  and  segments  coincide; 
whence,  arc  AwB  =  arc  AmB,  and  segment 
AnB  =  segment  AwB.     Q.  e.  d. 


Fig.  62. 


PROPOSITION    II. 

145.   Theorem. — The  diameter  of  a  circle  is  greater 
than  any  other  chord  of  the  same  circle. 

Demonstration. 

Let  AB  (Fig.  63)  be  a  chord  meeting  the  circumrerence  in  A  and  B, 
and  not  passing  through  the  centre  0  ;  and  let  AC  be  the  diameter. 


STRAIGHT    LINES    AND    CIRCUMFERENCES, 


69 


We  are  to  prove  that  AB  is  less  than  any 
diameter,  as  AC  (28). 

Now  as  AB  is  not  a  diameter,  it  does  not 
pass  through  0,  or  lie  in  AC  Hence  B  is  a 
different  point  from  C. 

Draw  OB. 

Now  AB  being  a  straight  line,  is  less  than 
A0  + OB,  which  is  a  broken  line  (69);  hence, 
as  AO  +  OB  =  AC,  AB  <  AC.     Q.  e.  d. 


146.  An  arc  is  said  to  be  Siibteiicled  by  the  chord  which 
joins  its  extremities,  and  the  arc  is  said  to  subtend  the  angle  in- 
cluded by  the  radii  drawn  to  its  extremities. 


PROPOSITION    III. 

147.  Theorem. — ^  radius  which  is  perpendicidar  to 
a  chord  bisects  the  chord,  the  subtended  arc,  and  the  sub- 
tended angle* 

Demonstration. 

Let  AB  be  a  chord  subtending  the  arc  AB,  which  arc  subtends  the 
angle  AOB.    Let  the  radius  EO  be  perpendicular  to  AB,  cutting  it  in  D. 

We  are  to  prove  that  DA  =  DB,  arc  AE 
=  arc  EB,  and  angle  AOE  =  angle  BOE. 

Produce  EO,  forming  the  diameter  EC. 

Revolve  the  semicircle  EBC  on  EC  as  an 
axis,  till  it  falls  in  the  plane  on  the  other  side 
of  EC. 

The  semicircles  will  coincide  (144),  and 
since  AB  is  perpendicular  to  EO,  DB  will  fall 

in  DA. 

Moreover,  as  OA  =  OB,  and  there  cannot  Fig.  64. 


*  Such  statements  in  Plane  Geometry  are  generally  limited  to  the  con- 
sideration of  arcs  less  than  a  semi  circumference,  yet  all  the  propositions 
in  this  section,  except  Prop.  VIII,  are  equally  true  whatever  the  arcs 
may  be. 


70 


ELEMENTARY    GEOMETRY. 


be  two  equal  oblique  lines  from  a  point  to  a  line  on  the  same  side  of  a 
perpendicular  (114),  OB  falls  in  OA,  and  B  falls  in  A. 

Hence,  DB  coincides  with  DA,  EB  with  EA.  and  angle  BOE  with  angle 
AOE,  and  we  have  DA  =  DB,  arc  AE  =  arc  EB,  and  angle  AOE  =  angle 

BOE.      Q.  E.  D. 


PROPOSITION    IV. 

148.  Theorem, — Conversely  to  Proposition  Til,  A  radius 
ivhich  bisects  an  arc  bisects  the  chord  which  subtends  the 
arCi,  is  perpendicular  to  the  chord,  and  also  bisects  the 
subtended  angle. 

Demonstration. 

Let  arc  AB  be  bisected  by  the  radius  OE  at  E.  Let  the  straight  line 
AB  be  the  chord  of  this  arc,  and  AOB  the  subtended  angle. 

We  are  to  prove  that  OE  bisects  the  chord 
AB  and  is  perpendicular  to  it,  and  also  bisects 
the  angle  AOB. 

Produce  EO,  forming  the  diameter  EC. 

Revolve  the  semi-circumference  EBC  upon 
EC  as  an  axis,  till  it  falls  in  the  plane  at  the 
left  of  EC. 

Then  will  semi  circumference  EBC  coincide 
with  EAC,  and  since  arc  BE  =  arc  AE  by  hy- 
pothesis, B  will  fall  in  A,  and  BD  =  AD.  Fig.  64. 

Hence,  the  line  OE  has  two  points,  0  and  D,  each  equally  distant 
from  A  and  B,  and  is  therefore  perpendicular  to  AB  (98)- 

Furthermore,   angle  BOD  coincides   with   AOD,  and   BOD  =  AOD- 

Q.  E.  D. 


^ 


PROPOSITION    V. 


149.  Theorem.— Conversely  to  Propositions  III  and  IV, 
A  radius  which  bisects  the  angle  included  by  two  other 
radii  bisects  the  arc  subtending  the  angle,  and  the  chord 
of  the  arc,  and  is  perpendicular  to  the  chord. 

[Let  the  student  give  the  demonstration.] 


STRAIGHT    LINES    AND    CIRCUMFERENCES. 


PROPOSITION     VI. 

150.  Theorem. — In  the  same  circle,  or  in  equal  cU'- 
ales,  equal  chords  are  equally  distant  from  the  centre. 

Demonstration. 

Let  EF  and  GH  be  equal  chords  in  the  same  circle  or  in  equal  circles ; 
and  OL  and  ON  be  the  perpendiculars  from  the  centre  0  upon  the 
chords,  and  thus  be  the  distances  of  the  chords  from  the  centre  (95). 


Fig.  65. 

We  are  to  prove  OL  =  ON. 

Since  OL  and  ON  are  perpendiculars  from  the  centre  upon  the  equal 
chords  EF  and  GH,  HN  =  FL  (147). 

Now  apply  the  figure  HNO  to  FLO,  placing  HN  in  its  equal  FL.  Then 
will  NO  coincide  with  LO  (88)- 

In  this  position,  HO  and  FO  are  equal  lines  drawn  from  the  same 
point  in  the  perpendicular  FL  to  the  line  LO-     Hence,  LO  =  NO  (116). 

Q.  E.  D. 

[Let  the  student  state  and  prove  other  converses  to  Propositions  III. 
IV  and  VI.] 


PROPOSITION    VII. 

151.  Theorem. — In  the  same  circle,  or  in  equal  cir- 
cles, if  two  arcs  are  equal,  the  chords  ivhich  subtend  them 
are  equal;  and,  conversely,  if  two  chords  are  equal,  the 
subtended  arcs  are  equal. 


72 


ELEMENTARY    GEOMETRY. 


Demonstration. 

Let  AaitB  and  Cfi^D  be  equal  arcs  in  the  same  circle  or  in  equal 
circles. 


Fig.  66. 


"We  are  to  prove,  first,  that  the  chords  AB  and  CD  are  equal. 

Apply  the  figure  C/iDO  to  AmBO,  placing  the  radius  CO  in  its  equal 
AO,  and  let  the  arc  CD  extend  in  the  direction  of  arc  AB. 

Then,  since  each  point  in  arc  CD  is  at  the  same  distance  from  the  cen- 
tre as  each  point  in  arc  AB,  arc  CD  falls  in  arc  AB,  and  since  arc  CD  = 
arc  AB  by  hypothesis,  D  falls  in  B. 

Hence,  chord  AB  =  chord  CD  (61).     Q.  e.  d. 

Conversely,  If  chord  AB  =  chord  CD,  arc  AB  =  aro  CD. 

Draw  the  perpendiculars  OL  and  ON.  Then,  since  the  chords  are 
equal,  OL  =  ON  (150). 

Now  apply  the  figure  CnDO  to  At/iBO,  placing  ON  in  its  equal  OL. 
Since  CD  is  perpendicular  to  ON,  and  AB  to  OL,  CD  will  fall  in  AB ;  and, 
since  the  chords  are  equal  by  hypothesis,  and  are  bisected  at  N  and  L 
(147),  D  falls  in  B  and  C  in  A. 

Hence,  arc  CwD  coincides  with  arc  AmB,  and  arc  CwD  =  arc  AwB. 

Q.  E.  D. 


PROPOSITION    VIII 


152.  Theorem.— /^i'  the  same  circle,  or  in  equal  cir- 
cles, if  two  arcs  are  unequal,  the  less  arc  has  the  less  chord ; 
and,  conversely,  if  two  chords  are  unequal,  the  less  chord> 
subtends  the  less  arc. 


straight  lines  and  circumferences,  73 

Demonstration. 
In  the  same  circle,  or  in  equal  circles,  let  arc  AmB  <  arc  CtiD 


Fig.  67. 

We  are  to  prove,  first,  that  chord  AB  <  chord  CD. 
Draw  OA,  OB,  OD,  and  OC 

Apply  the  figure  CnDO  to  AmBO,  placing  OC  in  its  equal  OA,  and  the 
arc  n  in  the  arc  m. 

Since  arc  CnD  >  arc  AwB,  D  will  fall  beyond  B,  as  at  D'.     Draw  OD'. 
AD'  will  evidently  cut  OB.     Let  N  be  the  point  of  intersection. 

Now  AB  <  AN  +  NB  (59), 

and  BO  =  D'O  <  ND'  +  ON  (59). 

Adding,  AB  +  BO  <  AN  +  ND'  +  ON  +  NB, 

or  AB  +  BO  <  AD'  +  BO. 

Subtracting  BO  from  each  member,  we  have 
AB  <  AD'.     Q.  E.  D. 

Conversely,  if  chord  AB  <  chord  CD. 

We  are  to  prove  that  arc  AmB  is  less  than  arc  CnD. 

For,  if  arc  AwiB  =  arc  C/iD,  chord  AB  =  chord  CD  (151).  And,  if 
arc  AwB  >  arc  C/iD,  chord  AB  >  chord  CD,  by  the  former  part  of  this 
demonstration.  But  both  of  these  conclusions  are  contrary  to  the  hy- 
pothesis. 

Hence,  as  arc  AmB  can  neither  be  equal  to  nor  greater  than  arc  CwD, 
it  must  be  less.     q.  e.  d. 


74  ELEMENTARY    GEOMETRY. 


PROPOSITION    IX. 

163.  Theorem. — In  the  same  circle,  or'  in  equal  civ- 
cles,  of  two  unequal  chords,  the  less  is  at  the  greater  dis- 
tance from  the  centre;  and,  conversely,  of  two  chords 
which  are  unequally  distant  from  the  centre,  that  which 
is  at  the  greater  distance  is  the  less. 

Demonstration. 

In  the  same  circle,  or  In  equal  circles,  let  chord  CE  <  chord  AB, 
and  OD  and  OD'  their  respective  distances  from  the  centre. 


Fig   68. 

We  are  to  prove,  first,  that  OD  >  OD'. 

From  A,  one  extremity  of  the  greater  chord,  lay  oflF  towards  B,  AE'  = 
CE.  Since  AE'  <  AB,  arc  AE'  <  arc  AB  (152),  and  E'  falls  somewhere 
on  the  arc  AB  between  A  and  B. 

Draw  OD"  perpendicular  to  AE',  and  OD"  =  OD,  since  the  equal 
chords  are  equally  distant  from  the  centre  (160). 

Now  OD"  is  a  different  line  from  OD',  since  OD"  produced  would 
bisect  arc  AE',  and  OD'  would  bisect  arc  AB.  Hence,  as  OD'  is  perpen- 
dicular to  AB,  OD"  must  be  oblique  (93). 

Again,  OD"  cuts  the  line  AB  in  some  point  as  H,  since  the  chord  AE' 
lies  on  the  opposite  side  of  AB  from  the  centre  0. 

Hence,  OH  >  OD'  (94),  and  much  more  is  OH  +  HD"  (=  OD")  >  OD'. 

Q.  E.  D. 

Conversely,  let  OD  >  OD'. 

We  are  to  prove  that  CE  <  AB. 

If  CE  =  AB,  OD  =r  OD'  (160),  and  if  CE  >  AB,  OD  <  OD',  both 
of  which  conclusions  are  contrary  to  the  hypothesis  OD  >  CD'. 

Hence,  as  CE  can  neither  be  equal  to  nor  greater  than  AB,  it  must  be 
less.     q.  E.  D. 


STRAIGHT    LINES    AND    CIRCUMFERENCES.  75 


PROPOSITION    X. 

154.  Theorem. — A  straight  line  which  intersects  a 
circumference  in  one  point  intersects  it  also  in  a  second 
point,  and  can  intersect  it  in  hut  two  points. 

Demonstration. 

Let  LM  (Fig.  69)  intersect  the  circumference  in  A. 

We  are  to  prove  that  it  intersects  in  finother  point,  as  B,  and  in  only 
these  two  points. 

Since  LM  intersects  the  circumference  in  A, I 
it  passes  within  it,  and  hence  has  points  nearerj 
the    centre   0  than   A.     OA  is,   therefore,   anf 
oblique  line,  and  not  the  perpendicular  from  0 
upon  LM  (94). 

Now  two  equal  oblique  lines  can  be  drawn! 
to  a  line  from  a  point  without  (114).  Let  OBI 
be  the  other  oblique  line  equal  to  OA.  But  asl 
OA  is  a  radius,  OB  =  OA  must  also  be  a  radiu3,[ 
and  B  is  in  the  circumference.     Q.  e.  d.  '^" 

Again,  LM  cannot  have  another  point  common  with  the  circumfer- 
ence, since  if  it  had  there  could  be  more  than  two  equal  straight  lines 
drawn  from  0  to  LM,  which  is  impossible.     Q.  e.  d. 

155.  Corollary. — Any  line  which  is  oblique  to  a  radius 
at  its  extremity  is  a  secant  line,  since  any  such  line  has  points 
nearer  the  centre  than  the  extremity  of  the  radius,  and  hence 
passes  within  the  circumference. 


PROPOSITION    XI. 

156.  Theorem. — A  straight  line  which  is  perpendicu- 
lar to  a  radius  at  its  outer  extremity  is  tangent  to  the 
circumference  ;  and,  conversely,  a  tangent  to  a  circumfer- 
ence is  perpendicular  to  a  radius  drawn  to  the  point  of 
contact. 

Demonstration. 

A  line  perpendicular  to  a  radius  at  its  extremity  touches  the  circum- 
ference because  the  extremity  of  the   radius  is  in  the  circumference. 


76 


ELEMENTARY    GEOMETRY. 


Moreover,  it  does  not  intersect  the  circumference,  since,  if  it  did,  it  would 
have  points  nearer  the  centre  than  the  extremity  of  the  radius ;  but  these 
it  cannot  have,  as  the  perpendicular  is  the  shortest  distance  from  a  point 
to  a  line.  Hence,  as  a  line  which  is  perpendicular  to  a  radius  at  its  ex- 
tremity touches  the  circumference  but  does  not  intersect  it,  it  is  a  tan- 
gent (30).     Q.  E.  D. 

Conversely,  as  a  tangent  to  a  circumference  does  not  pass  within,  the 
point  of  contact  is  the  nearest  point  to  the  centre,  and  hence  is  the  foot 
of  a  perpendicular  from  the  centre,     q.  e.  d. 

157.  Corollary. — A  perpendicular  from  the  centre  of 
a  circle  to  a  tangent  meets  the  tangent  in  the  point  of  tan- 
gencij  (93). 


PROPOSITION    XII. 

158.  Theorem. — The  arcs  of  a  circumference  inter- 
cepted by  two  parallels  are  equal. 

Demonstration. 

There  may  be  three  cases,  1st.  When  one  parallel  Is  a  tangent  and 
the  other  a  secant,  as  AB  and  CD ; 

2d.  When  both  parallels  are  secants,  as  CD  and  EF;  and 
3d.  When  both  parallels  are  tangents,  as  AB  and  GH. 

In  the  first  case  we  are  to  prove 
Ml  =  MK;  in  the  second,  IL  =  KR; 
and  in  the  third,  MwiN  =  M;iN. 

Through  0  draw  MN  perpendicular 
to  one  of  the  parallels,  in  any  case,  and 
it  will  be  perpendicular  to  the  other 
also  (133);  anfl  as  a  perpendicular 
from  the  centre  upon  a  tangent  meets 
the  tangent  at  the  point  of  tangency 
(157),  M  and  N  are  points  of  tangency, 
and  MN  is  a  diameter. 

Now,  since  the  parallels  are  perpendicular  to  MN,  and  the  chords  IK 
and  LR  are  bisected  by  it,  if  we  fold  the  right-hand  portion  of  the  figure 
on  MN  as  an  axis  until  it  falls  in  the  plane  on  the  left  of  MN,  K  will  fall 
in  I,  and  R  in  L. 

Hence,  Ml  =  MK,  IL  =  KR,  and  MmN  =  MwN.     Q.  e.  d. 


STRAIGHT   LINES    AND    CIRCUMFERENCES. 


77 


PROPOSITION    XIII. 
159.   Problem. — To  bisect  a  given  arc. 

Solution. 

Let  ACB  be  the  given  arc. 

We  are  to  bisect  it ;  that  is,  find  its  mid- 
dle point. 

Draw  the  chord  AB  joining  the  extremi- 
ties of  the  arc ;  and  bisect  this  chord  by  the 
perpendicular  00'  (101).  Then  will  00'  bi- 
sect the  arc,  as  at  C. 

Demonstration  of  Solution. 


Fig.  71. 


00'  being  a  perpendicular  to  the  chord  AB  at  its  middle  point,  any 
point  in  it  is  equally  distant  from  the  extremities.  Hence  chord  BO  - 
chord  AC,  and  arc  BC  =  arc  AC  (151).    Q-  e.  d. 


PROPOSITION    XIV. 

160.  Problem, — To  find  the  centre  of  a  circle  whose 
circumference  is  known,  or  of  any  arc  of  it. 

Solution. 
Let  ACB  be  an  arc  of  a  circumference. 

We  are  to  find  the  centre  of  the 
circle. 

Draw  any  two  chords  of  the  arc,  as 
AC  and  BC,  not  parallel,  and  bisect  each 
by  a  perpendicular.  Then  will  the  in- 
tersection of  these  perpendiculars,  as  0, 
be  the  centre  of  the  circle.  ^^^^^ 

Fig.  72. 

Demonstration  op  Solution. 

OL  being  perpendicular  to  the  chord  AC  at  its  centre,  passes  through 
the  centre  of  the  circle,  since  if  the  centre  were  out  of  OL  it  would 
be  unequally  distant  from  A  and  C  (96).    And  for  a  similar  reason,  CM 


78 


ELEMENTARY    GEOMETRY. 


being  perpendicular  to  the  chord  BC  at  its  centre,  passes  through  the 
centre  of  the  circle. 

Hence,  as  the  centre  of  the  circle  lies  at  the  same  time  in  LO  and  MO, 
it  is  their  intersection  0.    Q.  e.  d. 


PROPOSITION    XV. 

161.   Problem. — To    pass    a    circumference    through 
three  given  points  not  in  the  same  straight  line. 

SOLITTION. 

Let  A,  B,  and  C  be  the  three  given  points  not  in  the  same  straight 
line. 

Join  AB  and  BC. 

Bisect  AB  by  the  perpendicular  IVINj 
(101),  and  BC  by  the  perpendicular  RS. 

With  0,  the  intersection  of  MN  and  RS,| 
as  a  centre,  and  any  one  of  the  distances  0A,| 
OB,  OC,  say  OA,  as  a  radius,  describe  a  cir- 
cumference. 

Then  will  this  circumference  pass] 
through  the  three  points  A,  B,  and  C. 


Fig.  73. 


Demonstration  of  Solution. 

Since  AB  and  BC  are  non-parallel  by  hypothesis,  MN  and  RS  are  non- 
parallel  (141),  and  hence  meet  in  some  point,  as  0  (131). 

Now  as  every  point  in  MN  is  equally  distant  from  the  extremities  of 
AB  (96),  OA  =  OB. 

In  like  manner,  every  point  in  RS  is  equally  distant  from  B  and  C. 
Hence,  OB  =  OC. 

Hence,  OA  =  OB  —  OC,  and  a  circumference  struck   from  0  as  a 
centre,  with  a  radius  OA,  will  pass  through  A,  B,  and  C.     Q.  e.  d. 


PROPOSITION    XVI. 

162.  Problem. — To  draw  a  tangent  to  a  circle  at  a 
given  point  in  its  circumference. 


STRAIGHT   LINES    AND    CIRCUMFERENCES,  79 


Solution. 


Let  it  be  required  to  draw  a  tangent  to' 
the  circle  whose  centre  is  0,  at  the  point  P 
in  its  circumference. 

Draw  the  radius  OP,  and  produce  it  to 
any  convenient  distance  beyond  the  circle. 
Through  P  draw  MT  perpendicular  to  OP. 
Then  is  MT  a  tangent  to  the  circle  at  P. 


Demonstration  of  Solution. 


Fig.  74. 


MT  being  a  perpendicular  to  the  radius  at  its  extremity,  is  a  tangent 
to  the  circle  by  (166).    Q-  b.  d. 


EXE  RCISES. 

163.  1.  Draw  a  circle  and  divide  it  into  two  equal  parts. 
What  proposition  is  involved  ? 

2.  Given  a  point  in  a  circumference,  to  find  where  a  semi- 
circumference  reckoned  from  this  point  terminates.  Wliat 
proposition  is  involved  ? 

3.  In  a  circle  whose  radius  is  11  there  are  drawn  two  chords, 
one  at  6  from  the  centre,  and  one  at  4.  Which  chord  is  the 
greater  ?     By  what  proposition  ? 

4.  In  a  certain  circle  there  are  two  chords,  each  15  inches  in 
length.  What  are  their  relative  distances  from  the  centre? 
Quote  the  principle. 

5.  There  is  a  circular  plat  of  ground  whose  diameter  is  20 
rods.  A  straight  path  in  passing  runs  within  7  rods  of  the  cen- 
tre. What  is  the  position  of  the  path  with  reference  to  the  plat  ? 
What  is  the  position  of  a  straight  path  whose  nearest  point  is 
10  rods  from  the  centre?  One  whose  nearest  point  is  11  rods 
from  the  centre  ? 

6.  Pass  a  line  through  a  given  point,  and  parallel  to  a  given 
line,  by  the  principles  contained  in  (151),  (147),  (148),  and  (127). 


80 


ELEMENTARY    GEOMETRY, 


OF  THE  RELATIVE  POSITIONS  OF  CIRCUMFERENCES. 


AXI  OMS. 


164.  Two  circles  may  occupy  any  one  of  five  positions  with 
reference  to  each  other : 

1st.  One  circle  may  be  wholly  exterior  to  the  other. 

2d.  One  circle  may  be  tangent  to  the  other  externally,  the 
circles  being  exterior  to  each  other. 

3d.  One  circumference  may  intersect  the  other. 

4th.  One  circle  may  be  tangent  to  the  other  internally. 

5th.  One  circle  may  be  wholly  interior  to  the  other. 


PROPOSITION    I. 

165.  Theorem. —  WTien  one  circle  is  wholly  exterior  to 
another,  the  distance  between  their  centres  is  greater  than 
the  sum  of  their  radii. 

Demonstration. 

Let  M  and  N  be  two  circles 
whose  centres  are  0  and  0', 
and  whose  radii  are  OA  =  It, 
and  O'B  =  r,  respectively;  and 
let  N  be  wholly  exterior  to  M, 

We  are  to  prove  that  00' 
>  R  +  r. 

Draw  00',  and  let  it  inter- 
sect circumference  M  in  A,  and 
N  in  B. 


RELATIVE    POSITIONS    OF    CIRCUMFERENCES, 


81 


Since  N  is  wholly  exterior  to  M,  OB  >  OA. 

Adding  BO'  to  each  memlxjr  of  this  inequality,  we  have 
OB  +  BO'  >  OA  +  BO', 
or  00'  >  B  +  r, 

since  OB  +  BO'  =  00',  OA  =  i2,  and  OB  =  r.    q.  e.  d. 


PROPOSITION    II. 

166.  Theorem. —  When  two  circles  arc  tangent  to  each 
other  externally, 

1st  The  distance  between  their  centres  is  the  sum  of 
their  radii. 

2d.  They  have  a  common  rectilinear  tangent  at  their 
point  of  tangency. 

3d.  The  point  of  tangency  is  in  the  straight  line  join- 
ing their  centres. 

Demonstration. 

Let  M  and  N  be  two  circles  tangent  to  each  other  externally;  let  0 
and  0'  be  their  respective  centres,  B  and  r  their  radii,  D  the  point  of 
tangency,  and  TR  a  tangent  to  M  at  D. 


prove,  1st.  That 
2d.  That  TR  is 
3d.  That   D   is  in 


We  are    to 
00'  =  B  +  r; 

tangent  to   N ; 
00'. 

1st.  Draw  the  radii  OD  =  R, 
and  O'D  =  r. 

If  we  show  that  OD  +  O'D  = 
R  +  r  is  the  shortest  path  from  0 
to  0',  we  show  that  it  is  a  straight 
line  (59),  and  hence  is  the  distance 
from  0  to  0'  (95). 

Consider  any  other  path  from  0  to  0',  crossing  circumference  N  in 
some  other  point  than  D,  say  in  P. 

Now  the  shortest  path  from  0  to  P  is  the  straight  line  OP  (59) ;  and 
the  shortest  path  from  P  to  0'  is  the  straight  line  PO'.  Hence  the 
shortest  path  from  0  to  0'  passing  through  P  is  OP  -f  PO'. 


Fig.  76. 


82 


ELEMENTARY    GEOMETRY. 


But  OP  >  i?  (?)-*^,  and  PO'  =  r,  whence  OP  +  PO'  >  5  +  r. 

Hence,  as  P  is  the  point  where 
any  other  path  from  0  to  0'  crosses 
circumference  N,  OD  +  D'O  =  R 
+  r  is  the  distance  from  0  to  0'. 

Q.  E.  D. 

2d.  As  TR  is  tangent  to  M  at  D, 
by  hypothesis,  and  as  ODO'  has 
been  shown  to  be  a  straight  line,  TR 
is  perpendicular  to  DO'  (?)  and 
hence  tangent  to  N  (156).     Q-  e.  d. 

3d.  As  D  is  the  point  of  tangency,  and  ODO    is  00',  D  is  in  00'. 

Q.  E.  D. 


Fig.  76. 


PROPOSITION    III. 

167.  Theorem.— Tw/o  circumferences  which  intersect 
in  one  point  intersect  also  in  a  second  point,  and  hence 
have  a  cormnon  chord. 

DEMOi;rSTRATIOIT. 

Let  M  and  N  be  two  circumferences  intersecting  in  P. 


Fig.  77. 

We  are  to  prove  that  they  intersect  in  another  point,  as  P',  and  hence 
have  a  common  chord  PP'. 

As  M  intersects  N,  it  has  points  both  without  and  within  N. 

Now  consider  the  circumference  M  as  generated  by  a  point  moving 
from  left  to  right,  and  let  Y  be  a  point  within  N.     The  generating  point, 

*  Hereafter,  minor  references  to  principles  on  which  a  statement  de- 
pends will  be  omitted,  and  the  interrogation  mark  substituted.  This  indi- 
cates that  the  student  is  to  give  the  principle.  In  this  case.  P  is  without  M 
since  by  hypothesis  N  is  external  to  M. 


RELATIVE    POSITIONS    OF    CIRCUMFERENCES. 


83 


in  passing  from  Y,  a  point  within  N,  to  X,  any  point  in  the  circumference 
M  without  N,  must  cross  circumference  N  at  some  point,  as  P',  since  this 
is  a  closed  curve. 

Moreover,  this  second  point,  P',  is  a  different  point  from  P,  since  a 
circumference  of  a  circle  does  not  cut  itself,  or  become  tangent  to  itself. 

Hence,  if  circumference  M  cuts  circumference  N  in  P,  it  cuts  it  also  in 
a  second  point,  as  P'.     (j.  e.  d. 

Finally,  since  P  and  P'  are  common  to  both  circumferences,  the 
circles  M  and  N  have  a  common  chord  PP'.     q.  e.  d. 


PROPOSITION    IV 


168.    Theorem.  —  Wken  two  circumferences  intersect, 
1st.  The  line  joining  their  centres  is  perpendicular  to 

their  common  chord  at  its  middle  point. 

2d.  The  distance  between  their  centres  is  less  than  the 

sum,  of  their  radii  and  greater  than  their  difference. 

Demonstration. 

Let  M  and  N  be  two  circumferences  intersecting  at  P  and  P' ;  let  0 
and  0'  be  their  centres,  and  li  and  r  their  radii  respectively,  li  being 
equal  to  or  greater  than  r. 

We  are  to  prove,  1st.  That 
00'  is  perpendicular  to  PP'  at 
its  middle  point ;  and  2d.  That 
00'  <  R  -k-  r,  and  00'  > 
R-r. 

Draw  OP  and  O'P. 

Ist.  Since  0  is  equally  dis- 
tant from  P  and  P',  and  0'  is 
also  equally  distant  from  P  and 
P'  (?)  ,  00'  is  perpendicular  to 
PP'  at  its  middle  point  (98). 

Q.  E.  D. 

2d.  As  P  is  not  in  00',  00'  < 
Again,  00'  + 


OP  +  PO'  (?),  or  00'  < 
OP  >  OP, 


R  -^  r. 


or  00'  +  r  >  R\ 

whence,  subtracting  r  from  each  member, 

00'  >  iJ  -  r. 


Q.   E.  D. 


u 


ELEMENTARY    GEOMETRY, 


PROPOSITION    V. 

169.  Theorem. —  WTien  the  less  of  two  circles  is  tangent 
to  the  other  internally, 

1st.  They  have  a  common  rectilinear  tangent  at  the 
point  of  tangency. 

2d.  Their  centres  and  the  point  of  tangency  lie  in  the 
same  straight  line. 

3d.  The  distance  between  the  cerCtres  is  equal  to  the  dif- 
ference of  their  radii. 


Demonstration-. 

Let  M  and  N  be  two  circles  whose  centres  are  0  and  0'  respectively, 
N  being  less  than  M  and  tangent  to  it  internally;  let  H  and  r  be  their 
radii,  and  D  the  point  of  tangency. 

We  are  to  prove,  1st.  That  they  have 
a  coramon  rectilinear  tangent  at  D ;  3d. 
That  0,  0',  and  D  are  in  the  same  straight 
line ;  and  3d.  That  00'  =  i^  —  r. 

1st.  Draw  TR  tangent  to  IVI  at  D. 
Draw  also  O'D,  and  any  other  line  from  0' 
to  TR,  as  O'E. 

Now,  since  E  is  without  the  circle  IVI 
(?),  and  IVI  is  without  N  (?),  O'E  >  O'D, 
and  O'D  is  perpendicular  to  TR  (94). 

Hence,  TR  is  tangent  to  N  (156),  and 
is  therefore  a  common  tangent,     q.  e.  d. 

2d.  Since  both  OD  and  O'D  are  perpendicular  to  TR  at  D  (?),  OD 
and  O'D  coincide  (88),  and  0  and  0'  lie  in  the  same  straight  line  with  D. 

Q.  E.  D. 

3d.  Since  00'  and  D  are  in  the  same  straight  line,  and  0'  is  between 
0  and  D,  00'  =  OD  —  O'D ;  that  is,  00'  =  -B  —  r,    q.  e.  d. 


Fig.  79. 


RELATIVE    POSITIONS    OF    CIRCUMFERENCES.  85 


PROPOSITION     VI. 

170.  Theorem. —  WTien  the  less  of  two  circles  is  wholly 
interior  to  the  other,  the  distance  between  the  centres  is 
less  than  the  difference  of  their  radii. 

Demonstration. 

Let  M  and  N  be  two  circles  whose  centres  are  0  and  0',  and  whose 
radii  are  R  and  r  respectively,  and  let  N  be  wholly  within  M. 

We  are  to  prove  that  00'  <  22  —  r. 

Produce  00'  till  it  meets  both  circumferences 
on  the  same  side  of  0  that  0'  is ;  and  let  the  inter- 
sections with  N  and  M  respectively  be  D  and  E. 

Then,  as  0,  0',  D,  and  E  lie  in  order  in  the 
same  straight  line, 

OD  <  OE;  J-, 

and  subtracting  O'D  from  each,  and  noticing  that  00  —  O'D  =  00', 
that  OE  =  i?,  and  O'D  =  r,  we  have 

00'  <  B—r.     Q.  E.  D. 


171.  General  Scholium. — The  converse  of  each  of  Props.  1, 11,  IV, 
V,  and  VI  is  also  true.  Thus,  if  the  distance  between  the  centres  is 
greater  than  the  sum  of  the  radii,  the  circles  are  wholly  exterior  the  one 
to  the  other ;  since  if  they  occupied  any  one  of  the  other  four  possible 
positions,  the  distance  between  the  centres  would  be  equal  to  the  sum  of 
the  radii,  less  than  their  sum,  equal  to  their  diflference,  or  less  than  their 
diflference ;  any  one  of  which  conclusions  vrould  be  contrary  to  the  hy- 
pothesis. 

In  like  manner,  the  converse  of  any  one  of  the  five  propositions  may 
be  proved. 

This  method  of  proof  is  called  The  Reductio  ad  Absurdum,  and 
consists  in  showing  that  any  conclusion  other  than  the  one  stated  would 
lead  to  an  absurdity. 


ELEME^TAR  Y    GEOMETR  Y, 


PROPOSITION    VII. 

172.  Theorem. — All  the  circumferences  which  can  be 
passed  through  three  points  not  in  the  same  straight  line 
coincide,  and  are  one  and  the  same. 

Demonstration. 
Let  A,  B,  and  C  be  three  points  not  in  the  same  straight  line. 

We  are  to  prove  that  all  tb&  circnmferences 
which  can  be  passed  through  them  coincide,  and 
are  one  and  the  same  circumference. 

By  (161)  a  circumftrence  can  be  passed  through 
A,  B,  and  C. 

Now  every  point  equally  distant  from  A  and  B 
lies  in  FD,  a  perpendicular  to  AB  at  its  middle 
point  (?).  And,  in  like  manner,  every  point  equally 
distant  from  B  and  C  is  in  HE,  a  perpendicular  to 
BC  at  its  middle  point. 

But  the  two  straight  lines  FD  and  HE  can  intersect  in  only  one  point. 

Hence  all  circumferences  which  can  pass  through  A,  B,  and  C  have 
their  centre  in  0,  and  their  radius  OA,  and  therefore  they  constitute  one 
and  the  same  circumference,     q.  e.  d. 


173.  Cor.  1. — Through  any  three  points  not  in  the  sam^e 
straight  line  a  circumference  can  he  passed,  and  hut  one. 

174.  Definition. — A  circle  is  said  to  be  determined  when 
the  position  of  its  centre  and  the  length  of  its  radius  are  known. 

175.  Cor.  2. — Three  points  not  in  the  same  straight  line 
determ^ine  a  circle. 

176.  Cor.  3. — Two  circumferences  can  intersect  in  only 
two  points. 

For,  if  they  have  three  points  common,  they  coincide,  and  form  one 
and  the  same  circumference. 


RELATIVE    POSITIONS    OF    CIRCUMFERENCES, 


87 


EXERCISES. 

177.  1.  The  centres  of  two  circles  whose  radii  are  10  and  7, 
are  at  4  from  each  other.  What  is  the  relative  position  of  the 
circumferences  ?  What  if  the  distance  between  the  centres  is  17  ? 
What  if  20  ?    What  if  2  ?    What  if  0  ?     What  if  3  ? 

2.  Given  two  circles  0  and  0'  (Fig.  82),  to  draw  two  others, 
one  of  which  shall  be  tangent  to  these  externally,  and  to  the 
other  of  which  the  two  given  circles  shall  be  tangent  internally. 
Give  all  the  principles  involved  in  the  construction.  Give  other 
methods. 


Fig.  82. 


Fig.  83. 


3.  Given  two  circles  whose  radii  are  6  and  10,  and  the  dis- 
tance between  their  centres  20.  To  draw  a  third  circle  whose 
radius  shall  be  8,  and  which  shall  be  tangent  to  the  two  given 
circles.  Can  a  third  circle  whose  radius  is  2  be  drawn  tangent  to 
the  two  given  circles  ?  How  will  it  be  situated  ?  Can  one  be 
drawn  tangent  to  the  given  circles,  whose  radius  shall  be  1? 
Why? 

4.  With  a  given  radius,  draw  a  circumference  (Fig.  83)  which 
shall  pass  through  a  given  point  and  be  tangent  to  a  given  line. 


ELEMENTABr    GEOMETRY, 


^^^tvnn^  yi 


OF    THE    MEASUREMENT    OF    ANGLES. 


178.  Two  angles  are  Commensurable  when  there  is  a 
common  finite  angle  which  measures  each.  When  they  have  no 
such  common  measure,  they  are  Incommensurable. 

179.  An  Angle  at  the  Centre  is  an  angle  included  be- 
tween two  radii. 

180.  An  Inscribed  Angle  is  an  angle  whose  vertex  is  in 
a  circumference,  and  whose  sides  are  chords  of  that  circumfer- 
ence. 

181.  Angles  are  said  to  be  measured  by  arcs,  according  to  the 
principles  developed  in  the  following  f)ropositions. 


PROPOSITION    I. 

182.  Theorem. — In  the  same  circle,  or  in  eqnal  circles, 
two  angles  at  the  centre  are  in  the  same  ratio  as  the  arcs 
intercepted  between  their  sides. 

Demonstration. 
There  are  three  cases  : 

CASE    I. 
When  the  angles  are  equal. 

Let  angle  AOB  =  angle  DOE  (Fig.  84)  in  the  same  circle  or  in  equal 
circles. 


MEASUREMENT    OF   ANGLES, 


89 


We  are   to    prove   that 

AOB  _  arc  AB  * 
DOE      arc  DE 

Apply  tbe  angle  DOE 
to  the  angle  AOB,  placing 
the  radius  OD  in  its  equal 
OA.  By  reason  of  the 
equality  of  the  angles 
DOE  and  AOB,  OE  will 
fall  in  OB,  and  E  in  B  (?). 

Hence  DE  coincides  with  AB,  and 

arc  AB 
arc  DE 


Fig.  84. 


=  1. 


But,  by  hypothesis, 
Hence, 


AOB 
DOE 

AOB 
DOE 


=  1. 


arc  AB 
arc  DE 


(66).     Q.  E.  D. 


CASE   II. 

When  tJie  angles  are  commensurable. 

Let  AOB  and  DOE  be  two  commensurable  angles  at  the  centre  in  the 
same  circle^  or  in  equal  circles.  . 


Fig.  85. 


*  This  method  of  writing  a  proportion  is  ado^ited  in  this  book  as  the 
more  elegant,  and  as  it  appears  to  be  coming  into  exclusive  use.  The  above 
is  the  same  as 

AOB  :  DOE  ::  arc  AB  :  arc  DE 
and  is  to  be  read  in  the  same  manner. 


90 


ELEMENTARY     GEOMETRY, 


Fig.  85. 


We  are  to  prove  that 


AOB  _  arc  AB 
DOE  ~  arc  DE  * 


As  the  angles  are  commensurable  by  hyp()thesis,  let  m  be  their  com- 
mon measure,  and  let  it  be  contained  5  times  in  AOB  and  8  times  in  DOE. 

8o  that 

AOB      5 

DOE       8* 

Conceive  the  angle  AOB  divided  into  5  partial  angles,  each  equal  to 

m,  and  the  angle  DOE  divided  into  8  such  partial  angles. 

Now  as  these  partial  angles  are  equal,  their  intercepted  arcs  are  equal 

(?),  and  as  AB  contains  5  of  them,  and  DE  8, 

arc  AB  _  5 


Hence, 


arc  DE 
AOB 
DOE 


arc  AB 
arc  DE 


(?).    q.  E.  D. 


CASE    III. 

When  the  angles  are  incoinmensurahle. 

Let  AOB  and  DOE  (Fig.  86)  be  two  incommensurable  angles  at  the 
centre,  in  the  same  circle,  or  in  equal  circles. 

«r  .  1-  .         AOB       arc  AB 

We  are  to  prove  that        ^^  =  ^^^-^ ' 

If  the  ratio  ^  is  not  equal  to  the  ratio  ^^5^,  let  it  be  ^rea<^r  ; 


and  let 


in  which  DL  is  less  than  DE- 


AOB  _  arc  AB 
DOE  ~  arcDL' 


MEASUREMENT    OF    ANGLES. 


91 


Fig.  86. 

Draw  OL,  and  divide  AOB  into  equal  parts,  each  less  than  LOE. 
Apply  this  measure  to  DOE,  beginning  at  DO.  At  least  one  line  of  di- 
vision win  fall  between  OL  and  OE.     Let  this  be  OK. 

Now  AOB  and  DOK  are  commensurable ;  hence,  by  Case  II, 


but  by  hjrpothesis 


AOB 
DOK 

AOB 
DOE 


arc  AB 
arc  DK  * 

arc  AB 
arc  DL* 


-^.  .,.       AOB  ,      AOB         ,  arc  AB  ,      arcAB 
Dmdmg  DOK  ^y  DOE'  ^"^  ai^DK  ^^  ar^DL'  ^"  ^*^^ 


DOE 
DOK 


arc  DL 


arc  DK 
But  this  conclusion  is  absurd,  since 


DOE 
DOK 


>  1,      and 


arc  DL 
arcDK 


<  1. 


AOB 


Thus  we  show  that  the  ratio  ^-^  cannot  be  greater  than  the  ratio 

— =^p ;  and  in  a  similar  manner  we  may  show  that  -r^r^  cannot  be  less 
arc  DE  ^  DOE 


than 


arc  AB 
arc^E 


AOB  . 
DOE 


Hence,  as  '^^  is  neither  greater  nor  less  than  ^^^^^  ,  it  is  equal  to 


arc  AB 
arc  DE 


arc  AB         .        ,  AOB       arc  AB 

=r^,  and  we  have    =r;^=.  =      ^^f. 

arc  DE '  DOE       arc  DE 


Q.  E.  D. 


[For  other  methods  of  demonstrating  this  important  theorem,  see 
Appendix.] 


92  ELEMENTARY    GEOMETRY. 

188.  Out  of  the  truth  developed  in  this  proposition  grows  the 
method  of  representing  angles  by  degrees,  minutes,  and  seconds,  as  given 
in  Trigonometry  (Part  IV,  3-6)-  It  will  be  observed,  that  in  all  cases,  if 
arcs  be  struck  with  the  same  radius^  from  the  vertices  of  angles  as  centres, 
the  angles  bear  the  same  ratio  to  each  other  as  the  arcs  intercepted  by 
their  sides.  Hence  the  arc  is  mid  to  measure  the  angle.  Though  this  lan- 
guage is  convenient,  it  is  not  quite  natural ;  for  we  naturally  measure  a 
quantity  by  another  of  like  kind.  Thus,  distance  (length)  we  measure  by 
distance.,  as  when  we  say  a  line  is  10  inches  long.  The  line  is  length  ;  and 
its  measure,  an  inch,  is  length  also.  So,  likewise,  we  say  the  area  of  a 
field  is  4  acres:  the  quantity  measured  is  a  surface;  and  the  measure,  an 
acre,  is  a  surface  also.  Yet,  notwithstanding  the  artificiality  of  the 
method  of  measuring  angles  by  arcs,  instead  of  directly  by  angles,  it  is 
not  only  convenient  but  universally  used ;  and  the  student  should  know 
just  what  is  meant  by  it. 

189.  A  Degree  is  ^^  part  of  the  circumference  of  a  circle;  a 
Minute  is  ^^  of  a  degree,  and  a  Second  is  ^V  of  a  minute.  This  is  the 
primary  signification  of  these  terms.  But  as  any  angle  at  the  centre  sus- 
tains the  same  ratio  to  any  other  angle  at  the  centre  as  do  their  subtended 
arcs,  we  speak  of  an  angle  as  an  angle  of  so  many  degrees,  minutes, 
and  seconds.  Thus,  an  angle  of  45  degrees  (written  45°)  means  an  angle 
at  the  centre  45  times  as  large  as  one  which  subtends  ^^  of  the  circumfer- 
ence, or  half  as  large  as  one  which  subtends  90°  of  the  circumference. 

This  idea,  as  well  as  the  notation  °, ',  ",  for  degrees,  minutes,  and 
seconds,  has  already  been  made  familiar  in  Arithmetic. 

190.  As  the  vertex  of  any  angle  may  be  conceived  as  the  centre  of  a 
circle,  the  intercepted  arc  of  whose  circumference  measures  the  angle,  we 
speak  of  all  angles  in  the  same  manner  as  of  angles  at  the  centre.  Thus, 
a  right  angle  is  called  an  angle  of  90°.  one-half  a  right  angle  is  an  angle 
of  45°,  a  straight  angle  is  an  angle  of  180°,  and  the  sum  of  four  right 
angles,  being  measured  by  the  entire  circumference,  is  an  angle  of  360°,  etc. 


PROPOSITION    II. 

191.  Theorem. — An  inscribed  angle  is  measured  by 
half  the  arc  intercepted  between  its  sides. 

Demonstration. 
Let  APB  be  an  angle  inscribed  in  a  circle  whose  centre  is  0. 


MEASUREMENT    OF   ANOLES. 


J 


We  are  to  prove  that  tlie  angle  APB  is  measured  by  one-half  the 
arc  AB. 

There  are  three  cases :  1st.  When  the  centre  is  in  one  side ;  2d.  When 
the  centre  is  witliin  the  angle;  and  3d.   When  it  is  without. 


CASE    I. 
When  the  centre,  0,  is  in  one  side,  ds  PB, 

Draw  the  diameter  DC  parallel  to  AP. 
By  reason  of  the  parallels  AP  and  CD, 
arc  AC  =  arc  PD  (158) ; 
and,  since  COB  =  POD  (?), 

arc  CB  =  arc  PD  (?). 
Hence,        arc  AC  =  arc  CB, 
and  arc  CB  =  i  arc  AB.  ^,     „ 

*  Fig.  87. 

Again,  since  the  parallels  AP  and  DC  are  cut  by  the  transversal  PB, 
the  angles  APB  and  COB  are  equal  (126). 

But  COB  is  measured  by  arc  CB  (?).  Hence,  APB  is  measured  by 
arc  CB  =  ^  arc  AB.    Q.  e.  d. 


CASE    II. 
When  the  centre  is  within  the  angle. 

Draw  the  diameter  PC. 

Now  by  Case  I,  APC  is  measured  by  ^  arc  AC, 
and  CPB  by  ^  arc  CB.  Hence  the  sum  of  these 
angles,  or  APB,  is  measured  by  ^  arc  AC  -I-  ^  arc 
CB,  or  ^  arc  AB.     q.  e.  d. 

CASE    III. 
Wlien  the  centre  is  withont  the  angle. 

Draw  the  diameter  PC. 

By  Case  I,  APC  is  measured  by  ^  arc  AC,  and 
BPC  by  ^  arc  BC.  Hence,  APB,  which  is  APC  — 
BPC,  is  measured  by 

^  arc  KQ  —  ^  arc  BC 

or  \  arc  AB.     Q.  e.  d. 


Fig.  88. 


Fig.  89. 


94  ELEMENTARY    GEOMETRY, 

192.  Corollary. — In  the  same  circle  or  in  equal  circles, 
all  angles  inscribed  in  the  same  segment  or  in  equal  seg- 
ments intercept  equal  arcs,  and  are  consequently  equal. 
If  the  segment  is  less  than  a  semicircle,  the  angles  are 
obtuse;  if  a  semicircle,  right;  if  greater  than  a  semi- 
circle, acute. 


Fig.  90. 

Illustration. — In  each  separate  figure  the  angles  P  are  equal  to 
each  other,  for  they  are  each  measured  by  half  the  same  arc. 

In  0,  each  angle  P  is  acute,  being  measured  by  ^m,  which  is  less  than 
a  quarter  of  a  circumference. 

In  0',  each  angle  P  is  a  right  angle,  being  measured  by  ^m',  which  is 
a  quadrant  (quarter  of  a  circumference). 

In  0",  each  angle  P  is  obtuse,  being  measured  by  ^m",  which  is 
greater  than  a  quadrant. 


PROPOSITION    III. 

193.  Theorem.— Any  angle  formed  by  two  chords  in- 
tersecting in  a  circle  is  measured  by  one-half  the  sum  of 
the  arcs  intercepted  between  its  sides  and  the  sides  of  its 
vertical,  or  opposite,  angle. 

Demonstration. 

Let  the  chords  AB  and  CD  (Fig.  91)  intersect  in  P. 

We  are  to  prove  that  angle  APD  (=  angle  CPB  ?)  is  measured  by 

^  (arc  AD  +  arc  CB) ; 

and  that  angle  BPD  (=  angle  CPA  ?)  is  measured  by 

^  (arc  BD  +  arc  CA). 


MEASUnSMENT    OF   ANGLES.  95 

•I    Draw  CE  parallel  to  AB. 

Arc   AE  =  arc   CB   (?) ;   whence,  arc   ED  = 
arc  AD  +  arc  CB. 

Now  the  inscribed  angle  ECD   is  measured 
by  I  arc  ED  =  1^  (arc  AD  +  arc  CB). 

But  ECD  =  APD  (0 ;  hence,  APD  (=  CPB) 
is  measured  by  |^  (arc  AD  4-  arc  CB).     q.  e.  d. 

Finally,    that    ARC,   or  its    equal    BPD,  is 
measured   by   |  (AC  +  BD),  appears   from   the  f^'Q-  ^•• 

fact  that  the  sum  of  the  four  angles  about  P 

being  equal  to  four  right  angles,  is  measured  by  a  whole  circumference 
(190). 

But  APD  +  CPB  is  measured  by  AD  +  CB ;  whence  APC  +  BPD,  or 
2APC,  is  measured  by  the  whole  circumference  minus  (AD  +  CB);  that 
is,  by  AC  +  BD.    Hence  APC  is  measured  by  \  (AC  4-  BD).    q.  e.  d. 

194.  Scholium. — The  case  of  the  angle  included  between  two  chords 
passes  into  that  of  the  inscribed  angle  in  the  preceding  proposition,  by 
conceiving  AB  to  move  parallel  to  its  present  position  until  P  arrives  at 
C  and  BA  coincides  with  CE.  The  angle  APD  is  all  the  time  measured 
by  half  the  sum  of  the  intercepted  arcs;  but,  when  P  has  reached  C,  CB 
becomes  0,  and  APD  becomes  an  inscribed  angle  measured  by  half  its  in- 
tercepted arc. 

In  a  similar  manner  we  may  pass  to  the  case  of  an  angle  at  the  centre, 
by  supposing  P  to  move  toward  the  centre  All  the  time  APD  is  meas- 
ured by  ^(AD  +  CB);  but,  when  P  reaches  the  centre,  AD  =  CB,  and 
|(AD  +  CB)  =  ^  (2  AD)  =  AD ;  i.  e.,  an  angle  at  the  centre  is  measured 
by  its  intercepted  arc. 


PROPOSITION    IV. 

196.  Theorem. — An  angle  included  between  two  se- 
cants meeting  iidthout  the  circle  is  measured  by  one-half 
the  difference  of  the  intercepted  arcs. 

Demonstration. 

Let  APB  (Fig.  92)   be  an   angle  included  between  the  secants  PA 
and  PB ;  and  let  the  intersections  with  the  circumference  be  C  and  D. 


96 


ELEMENTARF    GEOMETRY. 


We  are  to  prove  that  APB  is  measured  by 
^  (arc  AB  —  arc  CD). 

Draw  CE  parallel  to  PB. 

Now  arc  CD  =  arc  EB  (?).  Hence,  arc  AE 
=  arc  AB  —  arc  CD. 

Again,  ACE  =  APB  (0- 

But  ACE  is  measured  by  |  arc  AE  (?). 
Hence  APB  is  measured  by 

^  arc  AE  =  ^  (arc  AB  —  arc  CD),     q.  e.  d. 

196.  Scholium. — This  case  passes  into  that 
of  an  inscribed  angle,  by  conceiving  P  to  move  Fig.  92. 

toward  C,  thus  diminishing  the  arc  CD.     When 

P  reaches  C,  the  angle  becomes  inscribed ;  and  as  CD  is  then  0,  ^  (AB  — 
CD)  =  jr  AB.  Also,  by  conceiving  P  to  continue  to  move  along  PA,  CD 
will  reappear  on  the  other  side  of  PA,  hence  will  change  its  sign,*  and 
|(AE  —  CD)  will  become  ^  (AE  +  CD),  as  it  should,  since  the  anglejs 
then  formed  by  two  chords  intersecting  within  the  circumference. 


PROPOSITION    V. 

197.  Theorem. — ^11  equal  angles  whose  sides  inter- 
cept a  given  line,  and  wJiose  vertices  lie  on  the  same  side 
of  that  line,  are  inscribed  in  the  same  segment  of  which 
the  intercepted  line  is  the  chord. 

Demonstration. 

Let  APB,  APB,  AP"B,  etc.,  be  any  number  of 
equal  angles  whose  sides  intercept  the  given  line  AB. 

We  are  to  prove  that  the  vertices  P,  P',  P",  etc., 
all  lie  in  the  same  arc  of  which  AB  is  the  chord. 

Through  one  of  the  vertices,  as  P,  and  A  and  B 
describe  a  circumference. 

Now  the  angle  APB  is  measured  by  \  the  arc 
AwB,  and  as  the  other  angles  are  equal  to  this,  they 
must  have  the  same  measure. 

*  In  accordance  with  the  law  of  positive  and  negative  quantities  as  used 
in  mathematics,  whenever  a  continuously  varying  quantity  is  conceived  as 
diminishing  till  it  reaches  0,  and  then  as  reappearing  by  the  same  law  of 
change,  it  must  change  its  sign. 


Fig.  93. 


MEASUREMENT    OF    AXGLES. 


97 


But  suppose  any  one  of  them,  as  P',  had  its  vertex  within  the 
segment.  It  would  then  be  an  angle  included  between  two  chords 
drawn  from  A  and  B,  and  hence  would  be  measured  by  ^AmB  plus  some 
arc  (193). 

If,  on  the  other  hand,  the  vertex  P'  was  without  the  segment,  the 
angle  would  be  an  angle  included  between  two  secants,  and  would  be 
measured  by  ^A^nB  less  some  arc  (195) 

Hence,  as  P'  can  lie  neither  without  nor  within  the  arc  APB,  it  lies  in 
it.      Q.  E.  D. 

198.  Corollary. — All  right  angles  whose  sides  inter- 
cept a  given  line  are  inscribed  in  a  semicircle  whose 
diameter  is  the  given  line. 


PROPOSITION    VI. 

199.  Theorem. — An  angle  included  between  a  tan- 
gent and  a  chord  drawn  from  the  point  of  tangency  is 
measured  by  one-half  the  intercepted,  arc. 

Demonstration. 


Let  TPA  be  an  angle  included  be- 
tween the  tangent  TM  and  the  chord 
PA. 

We  are  to  prove  that  TPA  is  mea- 
sured by  \  arc  PnA. 

Through  A  draw  the  chord  AD 
parallel  to  TM. 

Then  is  PAD  =  TPA  (?). 

Now  PAD  is  measured  by  \Pmli  (?). 

Whence  TPA  is  measured  by 
^PmD.     But  PwD  equals  PnA  (?). 

Hence  TPA  is  measured  by  ^PnA. 

Q.  E.  D. 


Fig.  94. 


Exercise. — Show  that  APM  is  measured  by  \  arc  AwP. 

Also,  observe  how  the  case  of  two  secants  (195),  passes  into  this. 


98 


ELEMENTARY    GEOMETRY. 


PROPOSITION    VII. 

200.  Theorem, — J_n  angle  included  between  two  tan- 
gents is  measured  by  one-half  the  difference  of  the  inter- 
cepted arcs. 

DEMONSTRATrON. 

Let  APB  be  an  angle  included  between 
the  two  tangents  PA  and  PB,  tangent  at 
C  and  D. 

We  are  to  prove  that  APB  is  measured 

by 

^  (arc  CmD  —  arc  CnD). 

Draw  the  chord  CE  parallel  to  PB. 

Now  arc  CnD  =  arc  EwD  (?). 

Whence    arc  CE  =  arc  CwD  —  arc  OnD, 

Again,        ACE  =  APB  (?). 

But  ACE  is  measured  by  \  arc  CE  =  | 
(arc  CwiD  —  arc  CnD).    Hence  APB  is  measured  by  ^  (arc  CwD  —  arc 
CwD).      Q.  E.  D. 

201.  Scholium. — The  case  of  two  secants  (195)  becomes  this  by  sup- 
posing the  secants  to  move  parallel  to  their  first  position  till  they  botk 
become  tangents. 


Fig.  95. 


PROPOSITION   VIII 


202.  Tlieorem.  —  An  angle 
included  between  a  secant  and  a 
tangent  is  measured  by  one-half 
the  difference  of  the  intercepted 
arcs. 

[Let  the  student  write  out  the  demon- 
stration in  form.] 


Fig.  96. 


MEASUREMENT    OF   ANGLES. 


99 


PROPOSITION    IX. 

203.  Problem, — From  a  given  point  in  a  given  line  to 
draw  a  line  luhich  shall  mahe  with  the  given  line  a  given 
angle. 

SOLUTIOiT. 
Let  A  be  the  given  point  in  the  given  line  AB,  and  0  the  given  angle. 

We  are  to  draw  from  A  a 
line  which  shall  make  with 
AB  an  angle  equal  to  0. 

From  0  as  a  centre,  with 
any  convenient  radius,  de- 
scribe an  arc,  as  a6,  measuring 
the  angle  0. 

From  A  as  a  centre,  with 
the  same  radius,  describe  an 

arc  en  cutting  AB  and  extend  p.     ^^ 

ing  on   that   side  of  AB  on 
which  the  angle  is  to  lie.     Let  this  arc  intersect  AB  in  c. 

From  c  as  a  centre,  with  a  radius  equal  to  the  chord  oJ,  describe  an 
arc  cutting  cw,  as  at  d. 

From  A  draw  a  line  through  <Z,  as  AC. 

Then  will  CAB  be  the  angle  required. 

Demonstration  of  Solution. 

Arc  ab  measures  angle  0  (?). 

Arc  cd  =  arc  ah  (?). 

Hence,  angle  CAB  =  angle  0  (?). 


PROPOSITION    X. 

204.   Problem. — Through  a  given  point  to  draw  a  par- 
allel to  a  given  line, 

SOLLTION. 
Let  P  (Fig.  98)  be  the  given  point,  and  AB  the  given  line. 


100 


ELEMENTARY    GEOMETRY. 


We  are  to  draw  a  line  through 
P  which  shall  be  parallel  to  AB. 

From  P  as  a  centre,  with  any 
i-adius  sufficiently  great,  strike  an 
arc  cutting  AB,  as  at  a,  and  ex- 
tending on  the  same  side  of  AB 
that  the  parallel  is  to  lie.     Let  the  arc  be  ac. 

From  a  as  a  centre,  with  the  same  radius,  pass  an  arc  through  P,  cut- 
ting AB  in  some  point,  as  &. 

With  the  chord  IP  as  a  radius  and  a  as  a  centre,  strike  an  arc  cutting 
«c,  as  in  0. 

Draw  a  line  through  0  and  P,  and  it  will  be  the  parallel  required. 


Demonstratioi?^  of  Solution. 

The  arcs  Oa  and  P6  are  arcs  of  circles  with  equal  radii,  and  have 
equal  chords,  and  are  hence  equal  arcs  (?). 

Tiie  angles  OPa  and  Pab  are  equal,  since  they  are  measured  by  the 
equal  arcs  Oa  and  P6  (?). 

Hence  the  transversal  Pa  cuts  the  two  lines  MN  and  AB,  making  the 
alternate  angles  MP«  and  BaP  equal.  Wherefore  MN  is  parallel  to  AB, 
and  as  it  passes  through  the  given  point  P,  it  is  the  parallel  required. 
Q.  E.  D. 


PROPOSITION    XI. 

205.   Problem. — From  a  point  without  a  circle  to  draw 
a  tangent  to  the  circle. 

Solution. 

Let  O  be  the  centre  and  OT  the  radius 
of  the  given  circle,  and  P  the  given  point. 

We  are  to  draw  from  P  a  tangent  to 
the  circle. 

Join  P  with  the  centre  0  by  a  straight 
line. 

On  the  line  OP  describe  a  circle  inter- 
secting the  given  circle  in  T  and  T'. 

Through  the  points  P  and  T,  P  and  T' 
draw  the  straight  lines  PM  and  PM'.    These  will  be  the  required  tangents. 


Fig.  99. 


MEASUREMENT    OF   ANGLES, 


101 


Demonstration  of  Solution. 

Drawing  OT,  the  angle  OTP  is  a  riglit  angle,  since  it  is  inscribed  in  a 
semicircle  (192). 

Hence  PM  is  a  tangent  to  the  circle,  as  it  is  a  perpendicular  to  a  radius 
at  its  extremity,  and  as  it  passes  through  P  it  fulfills  the  conditions  of  the 
problem. 

In  like  manner,  PM'  is  seen  to  be  a  tangent  passing  through  P,  and 
the  problem  has  two  solutions.     Q.  e.  d. 

206.  Corollary. — Through  any  point  mithout  a  circle 
two  tangents  may  he  drawn  to  the  circle. 


PROPOSITION    XII. 

207.  Problem. — On  a  given  line  to  construct  a  segment 
which  shall  contain  a  given  inscribed  angle. 

Solution. 

Let  AB  be  the  given  line  and  0  the  given  angle. 

We  are  to  construct  a  segment 
on  AB  which  shall  contain  the  0  as 
an  inscribed  angle. 

At  one  extremity  of  AB,  as  B, 
construct  an  angle  ABC  equal  to  0, 
and  on  the  side  of  AB  opposite  to 
that  on  which  the  segment  is  to  lie. 

Erect  a  perpendicular  to  CB  at 
B,  and  one  to  AB  at  its  middle  point 
E.    Let  F  be  the  intersection  of  these  Fig.  lOO. 

perpendiculars. 

With  FB  (or  FA)  as  a  radius,  describe  a  circle.  Then  will  AWw"B  be 
the  segment  required ;  and  any  angle  inscribed  in  this  segment,  as  AHB, 
will  be  equal  to  0. 

Demonstration  of  Solution. 

Since  CB  and  AB  are  non-parallel  lines,  perpendiculars  erected  to 
them  will  meet  in  some  point  as  F  (141)  131)- 


102 


ELEMENTARY    GEOMETRY, 


F  being  a  point  in  the  perpendicu- 
lar to  AB  at  its  middle  point  FA  = 
FB  (96),  and  a  circle  struck  with  FB 
as  a  radius  and  F  as  a  centre  will 
pass  through  A.  Moreover  CB  will 
be  a  tangent  to  this  circle,  since  it  is 
perpendicular  to  a  radius  at  its  ex- 
tremity (156). 

Now  0  =  ABC  by  construction, 
and  ABC  being  an  angle  included 
between  a  tangent  and  a  chord,  is 
measured  by  half  the  intercepted  arc  AwB  (?). 

But  any  angle  inscribed  in  the  segment  Am'm"B  is  measured  by  ^  arc 
AwiB  (0,  and  hence ijquals  ABC  =,0.    Q.  e.  d.  0  \    ^ 


cr.>L  L 


PROPOSITION    XIII 
208.  Problem. — To  bisect  a  given  angle. 

Solution. 

Let  BOA  be  the  given  angle. 

We  are  to  draw  a  line  dividing  BOA 
into  two  equal  angles. 

With  any  convenient  radius  and  0  as 
a  centre,  describe  an  arc  cutting  the  sides 
OB  and  OA  at  h  and  a. 

From  a  and  &  as  centres,  with  equal 
radii,  strike  arcs  cutting  in  some  point, 
as  P. 

Through  0  and  P  draw  a  straight  line. 

Then  is  the  angle  BOA  bisected  by  OP,  and  BOP  =  POA. 


Fig.  lOf. 


Demonstration  of  Solution. 

OP   being  perpendicular  to  the  chord  of  arc  ab  (?)  bisects  the  arc 
(147).     Hence  arc  &D  =  arc  aO. 

But     »°gl«BOP=''J^*°      Therefore,  BOP  =  POA.    <i.  e.  d. 
angle  POA      arc  aD 


MEASUREMENT    OF   ANGLES, 


103 


EXE  RCISES 


209.    1.  To  find  a  point  in  a  plane  having  given  its  distances 
from  two  known  points. 

When  are  there  two  sohitions? 
When  but  one  solution  ? 
When  no  solution  ? 

2.  In  Fig.  102  there  are  4  pairs  of  equal  angles.     Which  are 
they,  and  why  ? 

Show  that  COB  =  ABD  +  CDB. 
Show  that  DOB  =  ABC  +  DAB. 


Fig.  102.  Fig.  103. 

210.  Concentric  Circles  are  circles  which  have  a  com- 
mon centre. 

3.  Draw  two  concentric  circles  (Fig.  103),  such  that  the 
chords  of  the  outer  circle  which  are  tangent  to  the  inner  shall 
be  equal  to  the  diameter  of  the  inner. 

4.  From  a  point  out  of  a  given  straight  line  to  draw  a  line 
making  a  given  angle  with  the  first  line. 

5.  Prove  that  if  two  circles  are  concentric,  any  chord  of  the 
outer  which  is  tangent  to  the  inner  is  bisected  at  the  point  of 
contact. 

6.  Prove  that  if  D  and  B  (Fig.  104)  are  right  angles,  A  and  C 
are  supplementary. 


104 


ELEMENTARY    GEOMETRY. 


7.  Prove  that  if,  in  the  adjoining 
figure,  the  opposite  sides  AB  and  DC, 
and  AD  and  BC  be  produced  till  they 
meet,  the  lines  which  bisect  the  in- 
cluded angles  will  be  perpendicular  to 
each  other. 

8.  Draw  a  triangle,  and  then  draw 
a  circle  about  it  so  that  all  its  angles 
shall  be  inscribed ;  L  e.,  circumscribe 
a  circle  about  a  triangle.    (See  161.) 


Fig.  104. 


♦  <» 


^^^irCTfoJI  VII. 


OF    THE    ANGLES    OF    POLYGONS,   AND   THE    RELATION 
'    BETWEEN   THE  ANGLES  AND   SIDES. 


OF    TRIANGLES. 

211.  A  Plane  Triangle,  or  simply  a  Triangle,  is  a  plane 
figure  bounded  by  three  straight  lines. 

212.  With    respect   to   their  sides,    triangles    are    distin- 
guished as   Scalene,  Isosceles, 
and  Equilateral. 

A  Scalene  Triangle  is 

a  triangle  which  has  no   two  Fig.  los. 

sides  equal,  as  (1)  or  (2).  ^^^^^^ 

An  Isosceles    Triangle  is  a  triangle    HHRB 
which  has  two  of  its  sides  equal  to  each  other,    ^E^^^^k 

as  (3).  Fig.  106. 


TRIANGLES. 


105 


An  Equilateral  Triangle  is  a  triangle 
which  has  all  tliree  of  its  sides  equal  each  to  each, 
as  (4). 


213.  With  respect  to  their  angles,  triangles  ^'9  io7. 
are  distinguished  as  acute  angled,   right    angled,   and    obtuse 
angled. 

An  Acute  Angled  Triangle  is  a  triangle  all  of  whose 
angles  are  acute,  as  (4). 

A  Right  Angled  Triangle  is  a  triangle  one  of  whose 
angles  is  right,  as  (2). 

An  Obtuse  Angled  Triangle  is  a  triangle  one  of  whose 
angles  is  obtuse,  as  (1). 

214.  A  circle  Circumscribes  a  figure  when  all  the  angles 
of  the  latter  are  inscribed. 


PROPOSITION    I. 

215.   Theorem. — The  sum  of  the  three  angles  of  a  tri- 
angle is  two  right  angles. 

Demonstration. 

Let  ABC  be  any  triangle. 
We  are  to  prove  that 

A  +  B  +  C  =  2  right  angles. 

Circumscribe  a  circle  about  the  triangle  (161)- 

Then  the  angle  A  is  measured  by  ^  the  arc 
BaC  (0,  the  angle  B  by  |  the  arc  C&A,  and  the 
angle  C  by  ^  the  arc  AcB. 

Hence  the  sum  of  the  three  angles,  or  A  +  B  +  f^'9-  '08. 

C.  is  measured  by  ^  the  sum  of  BaC  +  C&A  4  AcB,  or  ^  the  circumference. 

But  a  semi-circumference  is  the  measure  of  two  right  angles  (190)- 
Hence  A  +  B  +  C  =  2  right  angles,     q.  e.  d. 


106  ELEMENTARY    GEOMETRY. 

216.  CoROLLAKY  1. — A  triangle  can  have  only  one  right 
angle,  or  one  obtuse  angle.     Why  ? 

217.  Corollary  2.— Two  angles  of  a  triangle,  or  their 
sum,  being  given,  the  third  may  be  found  by  subtracting 
this  sum  from  two  right  angles,  i.  e.,  any  angle  is  the 
supplement  of  the  sum  of  the  other  two. 

218.  Corollary  3. — The  sum>  of  the  two  acute  angles  of 
a  right-angled  triangle  is  equal  to  one  right  angle ;  i.  e., 
they  are  complements  of  each  other. 

219.  Corollary  4. — If  the  angles  of  a,  triangle  are 
equal  each  to  each,  any  one  is  one-third  of  two  right  an- 
gles, or  two-thirds  of  one  right  angle. 


PROPOSITION    II. 

220.  Theorem.— IT^e  sides  of  a  triangle  sustain  the 
same  general  relation  to  each  other  as  their  opposite  an- 
gles; that  is,  the  greatest  side  is  opposite  the  greatest 
angle,  the  second  greatest  side  opposite  the  second  grea^test 
angle,  and  the  least  side  opposite  the  least  angle. 

Demonstration. 

Let  ABC  be  any  triangle  having  the  angle  C  greater  than  B.  and  B 
greater  than  A. 

We  are  to  prove  that  AB  opposite  C  is  the 
greatest  side,  AC  opposite  B  the  next  greatest, 
and  BC  opposite  A  the  least. 

Circumscribe  a  circle  about  the  triangle  (161). 

If  the  triangle  is  acute-angled,  the  arc  meas- 
uring any  angle  is  less  than  a  quarter  of  a  circum- 
ference (191). 

Now  the  angle  C  being  greater  than  B,  the 
arc  c  is  greater  than  arc  &  (?).     Hence,  the  chord  Fig.  I09. 

AB  is  greater  than  the  chord  AC. 


TRIANGLES. 


107 


In  like  manner,  the  angle  B  being  greater  than  the  angle  A,  the  arc 
I  is  greater  thaa^  arc  a  (?).  Hence  the  chord  AC  is  greater  than  the 
chord  BC. 

If  the  triangle  has  one  right  angle,  as  C,  Fig.  110,  this  angle  is 
measured  by  ^  the  semi-circumference  AcB,  and  inscribed  in  the  semi- 
circumference  ACB.  Hence  the  order  of  magnitude  of  the  arcs  is  still 
c>l>a  (?),  and  of  the  sides  AB  >  AC  >  BC. 


Fig.  110. 


Fig.  III. 


If  any  angle  of  the  triangle,  as  C,  is  obtuse.  Fig.  Ill,  this  angle  is  in- 
scribed in  a  segment  less  than  a  semicircle  (192),  whence  this  arc  ACB  is 
less  than  a  semi-circumference,  and  greater  than  either  a  or  J,  as  it  is  their 
sum. 

Hence  the  chord  AB  is  greater  than  either  AC  or  BC  (?). 

Thus  we  have  shown  that  in  all  cases,  the  order  of  magnitude  of  the 
angles  being  C  >  B  >  A,  the  order  of  magnitude  of  the  sides  is 
AB  >  AC  >  BC.       Q.  E.  D. 


221.  Corollary  1. — Conversely,  TJie  order  of  the  Tnag- 
nitudes  of  the  sides  being  AB  >  AC  >  BC,  the  order  of  the 
magnitudes  of  the  angles  is  C>  B  >  A. 

[Let  the  student  give  the  demonstration  in 
form.] 

222.  Corollary  2. — An  equiangu- 
lar triangle  is  also  equilateral ;  and, 
conversely,  an  equilateral  triangle  is 
equiangular. 

Thus,  if  A  =  B  =  C,  arc  a  =  arc  6  =  arc  c,  p.     ,,2 


108 


ELEMENTARY    GEOMETRY. 


and,  consequently,  chord  BC  =  chord  AC  =  chord  AB.  Conversely,  if 
the  chords  are  equal,  the  arcs  are,  and  hence  the  angles  subtended  by 
these  arcs. 


223.  Corollary  3. — In  an  isosceles  triangle  the  angles 
opposite  the  equal  sides  are  ecfual ;  and, 
conversely,  if  two  angles  of  a  triangle 
are  equal,  the  sides  opposite  are  equal, 
and  the  triangle  is  isosceles. 

Thus,  if  AB  =  BC,  arc  a  =  arc  c ;  and  hence, 
angle  A,  measured  by  ^  a,  =  angle  C,  measured 
by|c. 

Conversely,  if  A  =  C,  arc  a  =  arc  c ;  and 
hence  chord  BC  =  chord  AB. 

224.  Scholium. — It  should  be  observed  that  the  proposition  gives 
only  the  general  relation  between  the  angles  and  sides  of  a  triangle.  It  is 
not  meant  that  the  sides  are  in  the  same  ratio 
as  their  opposite  angles:  this  is  not  true. 
Thus,  in  Fig.  114,  angle  C  is  twice  as  great  as 
angle  A  ;  but  side  c  is  not  twice  as  great  as  side 
«,  although  it  is  greater.  Trigonometry  dis- 
covers the  exact  relation  which  exists  between 
the  sides  and  angles. 


PROPOSITION    III. 

225.  Theorem. — //  from  any  point  ivithin  a  triangle 
lines  are  drawn  to  the  extremities  of  any  side,  the  included 
angle  is  greater  than  the  angle  of  the 
triangle  opposite  this  side. 

Demonstration. 

Let  ACB  be  any  triangle,  0  any  point  with- 
in, and  OB  and  DA  lines  drawn  from  this  point 
to  the  extremities  of  AB. 

We  are  to  prove  that  angle  AOB  >  angle 

ACB.  Fig.  115. 


TRIANGLES, 


109 


Circumscribe  a  circle  about  the  triangle  (161),  and  produce  AO  and 
BO  till  they  meet  the  circumference. 

Now  ACB  is  measured  by  ^AnB  (191);  but  AOB  is  measured  by 
|(AnB  +  EmD)  (193).     Hence^  AOB  >  ACB.     Q.  e.  d. 


226.  An  Exterior  Angle  of  a  triangle  is  an  angle  formed 
by  any  side  with  its  adjacent  side  produced,  as  CBD,  Fig.  116. 


PROPOSITION    IV. 

227.   Theorem.— v^/i'i/  exterior  angle  of  a  triangle  is 
equal  to  the  sum  of  the  two  interior  non-adjacent  angles. 


Demonstration. 

Let  ABC  be  a  triangle,  and  CBD  be^n  ex- 
terior angle. 

We  are  to  prove  that  CBD  =  A  +  C. 
ABC  +  CBD  =  a  straight  angle  (?). 
But  ABC  +  A  +  C  =  a  straight  angle  (?). 
Hence,  ABC  +  CBD  =  ABC  +  A  +  C  (?). 
Hence,  subtracting  ABC  from  each  member, 

CBD  =  A  +  C.    Q.  E.  D. 


Fig.  116. 


228.  Corollary. — Either  angle  of  a  triangle  not  adja- 
cent to  a  specified  exterior  angle,  is  cqiuil  to  the  differ- 
ence between  this  exterior  angle  and  the  other  non- 
adjacent  angle. 

Thus,  since  CBD  =  A  +  0, 

by  transposition,  CBD  —  A  =  C, 

and  CBD  —  C  =  A. 


110 


ELEMENTARY    GEOMETRY. 


OF    QUADRILATERALS. 

229.  A  Quadrilateral  is  a  plane  surface  inclosed  hj  four 
right  lines. 

230.  There  are  three  Classes  of  quadrilaterals,  viz.,  Trape- 
ziums, Trapezoids,  and  Parallelograms, 

231.  A  Trapezium  is  a  quadrilateral  which  has  no  two 
of  its  sides  parallel  to  each  other. 

232.  A  Trapezoid  is  a  quadrilateral  which  has  but  two  of 

its  sides  parallel  to  each  other. 

233.  A  Parallelogram  is  a  quadrilateral  which  has  its 
opposite  sides  parallel. 

234.  A  Rectangle  is  a  parallelogram  whose  angles  are 
right  angles. 

235.  A  Sqviare  is  an  equilateral  rectangle. 

236.  A  Rhombus  is  an  equilateral  parallelogram  whose 
angles  are  oblique. 

237.  A  Rhom- 
boid is  an  oblique- 
angled  parallelogram 
two  of  whose  sides 
are  greater  than  the 
other  two. 

III.  — The  figures  in 
the  margin  are  all  quad- 
rilaterals. A  is  a  trape- 
zium. (Why  ?)  B  is  a 
trapezoid.  (Why?)  C, 
D,  E,  and  F  are  paral- 
lelograms. (Why?)  D 
and   E  are  rectangles, 

Fig.  117. 


Q  UADRILA  TERALS. 


Ill 


although  D  is  the  form  usually  referred  to  by  the  term  rectangle.  So  C 
is  the  form  usually  referred  to  when  a  parallelogram  is  spoken  of,  without 
saying  what  kind  of  a  parallelogram.  C  is  also  a  rhomboid.  (Why  ?) 
E  is  a  square.     (Why  ?)     F  is  a  rhombus.     (Why?) 

238.  A  Diagonal  is  a  line  joining  the  vertices  of  two  non- 
consecutive  angles  of  a  figure. 

239.  The  Altitude  of  a  parallelogram  is  a  perpendicular 
between  its  opposite  sides  ;  of  a  trapezoid,  it  is  a  perpendicular 
between  its  parallel  sides ;  of  a  triangle,  it  is  the  perpendicular 
from  any  vertex  to  the  side  opposite  or  to  that  side  produced. 

240.  The  Bases  of  a  parallelogram,  or  of  a  trapezoid,  are 
the  sides  between  which  the  altitude  is  conceived  as  taken  ;  of  a 
triangle,  the  base  is  the  side  to  which  the  altitude  is  perpendiculai*. 


PROPOSITION    V. 

241.  Theorem. — ITie  sum  of  the  angles  of  a  quadri- 
lateral is  four  right  angles.* 


Demonstration. 

Let  ABCD  be  any  quadrilateral. 

We  are  to  prove  that 
DAB  +  B  +  BCD  +  D  =  4  right  angles. 

Draw  either  diagonal,  as  AC. 

The  diagonal  divides  the  quadrilateral 
into  two  triangles,  and  the  sum  of  the  an- 
gles of  the  two  triangles  is  the  same  as  the 
sum  of  the  angles  of  the  quadrilateral,  since 

BCA  +  ACD  =  BCD, 
and  BAC  +  CAD  =  DAB. 

But  the  sum  of  the  angles  of  the  triangles  is  four  right  angles  (?). 
Hence  the  sum  of  the  angles  of  the  quadrilateral  is  four  rii;ht  angles. 

<l.  B.D. 

*  See  (261). 


Fig.  118. 


112 


ELEMENTARY    GEOMETRY, 


PROPOSITION    VI. 

242.  Theorem. — The  opposite  angles  of  any  quadri- 
lateral which  can  be  inscribed  in  a  circle  are  suppLe- 
mentary. 

DEMOKSTRATIOiq^. 
Let  ABCD  be  any  inscribed  quadrilateral. 

We  are  to  prove  that 

A  +  C  =  2  right  angles, 
and  also  that    D  +  B  =  2  right  angles. 

A  is  measured  by  ^  the  arc  DCS,  and  C  by  ^ 
the  arc  BAD. 

Hence,  A  +  C  is  measured  by  ^(DCB  +  BAD), 
that  is,  by  a  semi-circumference,  and  is  therefore 
2  right  angles  (190).     Q-  e.  d. 


Fig.  119. 


In  like  manner,  B  +  D  is  measured  by  ^(ADC  +  CBA),  and  hence  is 
3  right  angles.     Q.  e.  d. 


PROPOSITION    VII. 

243.  Theorem. — The  adjacent  angles  of  a  parallelo- 
gram are  supplemental,  and  the  opposite  angles  are  equal 
to  each  other. 

r  Demokstration. 

Let  ABCD  be  any  parallelogram. 

We  are  to  prove,  1st.  That  A  +  B,  or 
B  +  C,  or  C  +  D  ,  or  D  +  A  is  2  right 
angles ;  and  2d.  That  A  =  C  and  D  =  B. 

1st.  Since,  by  definition  (233),  AD  is  ^'9-  '^^• 

parallel  to  BC,  and  the  transversal  AB  cuts  them,  the  sum  of  the  two  in- 
terior angles  on  the  same  side,  that  is,  A  +  B,  is  2  right  angles  (126). 

In  like  manner,  B  +  C  is  two  right  angles,  since  they  are  the  interior 
angles  on  the  same  side  of  the  transversal  BC  which  cuts  the  parallels  AB 
and  DC. 


QUADRILATERALS.  113 

In  the  same  way,  C  +  D,  or  D  +  A  maybe  shown  equal  to  2  right 

angles. 

Hence  the  sum  of  any  two  adjacent  angles  of  a  parallelogram  is  2 
right  angles,     q.  e.  d. 

2d.  A  +  B  =  B  +  C,  since  each  sum  is  2  right  angles,  by  the  pre- 
ceding part  of  this  demonstration. 

Hence,  subtracting  B  from  each  member,  we  have  A  =  C. 

In  a  similar  manner,  we  may  show  that  B  =  D. 

Hence,  either  two  opposite  angles  are  equal  to  each  other,     q.  k  d. 

244.  Corollary  1. — The  two  angles 
of  a  trapezoid  adjacent  to  either  one 
of  the  two  sides  not  parallel  are  sup- 
plemental. Fig.  121. 

[Let  the  student  show  why.] 

245.  Corollary  2. — //  one  angle  of  a  parallelogram  is 
right,  the  others  are  also,  and,  the  figure  is  a  rectangle. 


PROPOSITION    VIII. 

246.  Theorem. — Conversely  to  the  last,  If  three  consec- 
utive angles  of  a  quadrilateral  are  such  that  the  first 
and  the  second,  and  the  second,  and  the  third,  are  sup- 
plemental, or  if  the  opposite  angles  are  equal,  the  figure  is 
a  parallelogram. 

Demonstratiok. 

Let  ABCD  be  a  quadrilateral  having  D  and  A,  and  A  and  B  supple- 
mental, or  having  A  =  C  and  D  =  B. 

We  are  to  prove  that,  in  either  case, 
the  figure  is  a  parallelogram. 

1st.  If  we  have  D  and  A,  and  A  and  B 
supplemental.  ^.^  ,22. 

Since  the  transversal  AD  cuts  the  lines 
AB  and  DC,  making  A  +  D  =  2  right  angles,  the  lines  AS  and  DC  are 
parallel  (126). 


114 


ELEMENTARY    GEOMETRY, 


Again,  for  a  like  reason,  since  A  +  B  =  2  right  angles,  AD  and  BC  are 
parallel. 

Hence  the  opposite  sides  of  the  quad- 
rilateral are  jDarallel,  and  the  figure  is  a 
parallelogram  (23d)<    Q.  e.  d. 


Fig.   172. 


2d.  If  A  =  C,    and   D  =  B,    adding, 
we  have 

A  +  D  =  C  +  B. 

But  A  +  D-l-C  +  B  =  4  right  angles. 

Hence,  substituting,  we  have 

A  +  D  +  A  +  D  =  4  right  angles  (?), 
(Mr  2  (A  +  D)  =  4  right  angles, 

or  A  +  D  =  A  +  B  (?)  =  2  right  angles, 

and  the  figure  is  a  parallelogram  by  the  former  part  of  the  demonstration. 

Q.  B.  D. 


PROPOSITION    IX. 

247.    Theorem. — //  two  opposite  sides  of  a  quadrilat- 
eral are  equal  and  parallel,  the  figure  is  a  parallelogram, 

Demokstratiok. 
Let  ABCD,  in  (a),  be  a  quad- 
rilateral having  the  sides  AB  and 
DC  equal  and  parallel. 

We  are  to  prove  that  AD  and 
BC  are  parallel,  and  hence  that 
the  figure  is  a  parallelogram. 

Draw  the  diagonal  AC. 

Then,  by  reason  of  the  paral- 
lels AB  and  DC.  the  angles  BAC 
and  DCA  are  equal  (?) 

Conceive  the  quadrilateral  di- 
vided in  this  diagonal  into  two 
triangles,  as  in  (b). 

Reverse  the  triangle  ACB  and 
place  it  as  in  (c).  Since  AC  of 
the  triangle  ADC  =  CA  of  the 
triangle  ABC,  CA  may  be  placed  in  AC,  as  in  (c) 


Fig.  123. 


Q  UA  DRILA  TERAL8, 


115 


Now  revolve  the  triangle  CBA  on  CA  as  an  axis.  Since,  as  we  have 
shown,  the  angle  BAC  =  angle  DCA,  BA  will  take  the  direction  CD,  and 
being  equal  to  it,  by  hypothesis,  B  will  fall  in  D,  and  the  angle  BCA  co- 
incides with  and  is  equal  to  DAC. 

But  in  {a)  the  angles  BCA  and  DAC  are  alternate  interior  angles  made 
by  the  transversal  AC  cutting  AD  and  BC.  Hence  AD  and  BC  are  paral- 
lel, and  as  AB  and  DC  are  parallel  by  hypothesis,  the  quadrilateral  is  a 
parallelogram  (233).    Q-  e.  d. 


PROPOSITION    X. 

248.    Theorem. — //  the  opposite  sides  of  a  quadrilat- 
eral are  equal,  the  figure  is  a  parallelogram. 

Demonstration. 
Let  ABCD,  (a),  be  a  quadrilateral,  having  AD  =  BC  and  AB  =  DC. 

We  are  to  prove  that  ABCD 
is  a  parallelogram,  i.  e.,  that 
AB  is  parallel  to  DC,  and  AD  to 
BC. 

Draw  the  diagonal  AC,  and 
conceive  the  quadrilateral  di- 
vided in  this  diagonal  into  two 
triangles,  as  in  (&). 

Reverse  the  triangle  ABC, 
and  place  it  as  in  (c).  Since 
AC  of  the  triangle  ADC  equals 
CA  of  the  triangle  ABC,  CA 
may  be  placed  in  AC,  as  in  (c). 

Draw  DB,  intersecting  CA 
(or  CA  produced),  in  0. 

As  CD  =  AB,  and  AD  = 
CB,  by  hypothesis,  the  line  AC 
has  two  points  each  equally  distant  from  the  extremities  of  DB,  and  AC 
and  DB  are  perpendicular  to  each  other  (98)-  Moreover,  since  AB  and 
CD  are  equal  oblique  lines  drawn  from  the  same  point  in  the  perpendic- 
ular to  the  line  DB,  angle  BAC  =  angle  DCA  (98,  110,  2d). 

Now  in  (a),  as  angles  BAC  and  DCA  are  the  alternate  interior  angles 
made  by  the  transversal  with  the  lines  AB  and  DC,  the  latter  are  parallel, 


Fig.  124. 


116 


ELEMENTARY     GEOMETRY, 


and  as  they  are  equal  by  hypothesis,  the  quadrilateral  is  a  parallelogram 
by  the  last  proposition,     q.  e.  d. 

249.   Corollary. — A  diagonal  of  a  parallelogranv  di- 
vides it  into  two  equal  triangles. 


PROPOSITION    XI. 

250.    Theorem. — Conversely  to  the  last,   The  opposite 
sides  of  a  parallelogram  are  equal. 

Demonstration. 


Let  ABCD  be  a  parallelogram. 
We  are  to   prove  that  AD  =  BC, 
and  AS  =  DC. 

AD  and  BC  being  parallel  transver- 
sals cutting  the  parallels  AB  and  DC,  their 
intercepted  portions,  which  are  the  oppo- 
site sides  of  the  parallelogram,  are  equal  by  (138) 

For  a  like  reason,  AB  =  DC. 

Hence,  AD  =  BC  and  AB  =  DC.    Q.  e.  d. 


Fig.  125. 


PROPOSITION     XII. 

251.    Theorem. — The  diagonals  of  a  parallelogram 
bisect  each  other. 

Demonstration. 

Let  ABCD  be  a  parallelogram  whose 
diagonals  AC  and  DB  intersect  In  Q. 

We  are  to  prove  that  AQ  =  QC,  and 
DQ  =  QB. 

Angle   QDC  -  angle  QBA   (?),  angle 
QCD  =  angle  QAB  (?),  and  DC  =  AB  (?).  ''■g-  '26. 

For  distinctness,  let  Q'  represent  the  vertex  at  Q  of  the  triangle  DQC. 

Now  apply  the  triangle  AQB  to  DQ'C,  placing  the  side  BA  in  its  equal 
DC,  with  the  extremity  B  in  D,  and  A  in  C,  and  the  vertex  Q  on  the  same 


Q  UADRILA  TERALS. 


117 


side  of  DC  that  the  vertex  Q'  is,  and  the  triangles  will  coincide.  For, 
since  angle  QAB  =  angle  Q'CD,  AQ  will  take  the  direction  CQ',  and  the 
vertex  Q  will  fall  somewhere  in  the  line  CQ'.  In  like  manner,  by  reason 
of  the  equality  of  angles  QBA  and  Q'DC,  the  vertex  Q  will  fall  in  DQ'. 
Hence  the  vertex  Q  of  the  triangle  AQB  falling  at  the  same  time  in  CQ' 
and  DQ',  falls  at  their  intersection. 

Hence,  as  these  triangles  coincide,  AQ  =  Q'C,  and  DQ'  =  QB ;  that  is, 
AQ  =  QC,  and  DQ  =  QB.     q.  e.  d. 


PROPOSITION     XIII. 

252.  Theorem. — The  diagonals  of  a  rlLombus  bisect 
each  other  at  Hght  angles. 

Demonstration. 

Let  ABCD  be  a  rhombus,  and  AC  and  DB  its  diagonals  intersecting 
at  Q. 

We  are  to  prove  that  DB  and  AC  are  per- 
pendicular to  each  other. 

Since  AB  =  AD,  and  CD  =  CB  (?),  the 
line  AC  has  two  points,  A  and  C,  each  equally 
distant  from  the  extremities  of  DB.  Hence  AC 
is  a  perpendicular  to  DB  at  its  middle  point  Q 
(98).     Q.  E.  D.  P5g-  '27. 

In  like  manner,  DB  may  be  shown  to  be  perpendicular  to  AC  at  its 
middle  point.    Q.  e.  d. 

253.  Corollary. — The  diagonals  of  a  rhombus  bisect  its 

angles. 

For,  revolve  ABC  upon  AC  as  an  axis,  and  it  will  coincide  with  ADC. 
Hence  angles  A  and  C  are  bisected.  In  like  manner  revolve  DAB  upon 
DB,  and  it  will  coincide  with  DCB.     Hence,  D  and  B  are  bisected. 


PROPOSITION    XIV. 

254.   Theorem. — The  diagonals  of  a   rectangle  are 
equal. 


118  ELEMEyTARY    GEOMETBT. 

Demonstration. 

Let  AC  and  DB  be  the  diagonals  of  the  rectangle  ABCD, 

We  are  to  prove  that  AC  =  DB; 

Upon  AC  as  a  diameter  describe  a  circle. 

Since  ADC  and  ABC  are  right  angles  whose 
sides  intercept  AC,  they  are  inscribed  in  the  cir- 
cumference of  which  AC  is  a  diameter  (198). 

Again,  since  DCB  is  a  right  angle  and  is  in- 
scribed, DB  is  a  diameter  (?). 

Hence  AC  and  DB,  being  diameters  of  the  same 
circle,  are  equal,     q.  e.  d. 


Fig.  128. 


255.   Corollary. — Conversely,  //  the  diagoTials  of  a  par- 
allelogram are  equal,  the  figure  is  a  rectangle. 

By  (251)  the  parallelogram  is  circuniscriptible ;  whence,  by  (192)  the 
angles  are  right  angles. 


OF   POLYGONS   OF   MORE   THAN    FOUR 

SIDES. 

256.  A  Polygon  is  a  portion  of  a  plane  bounded  by  straight 
lines. 

The  word  polygon  means  many-angled  ;  so  that  with  strict  propriety 
we  might  limit  the  definition  to  plane  figures  with  five  or  more  sides. 
This  limitation  in  the  use  of  the  word  is  frequently  made. 

267.  A  polygon  of  three  sides  is  a  triangle  ;  of  four,  a  quad- 
rilateral ;  of  five,  a  pentagon;  of  six,  a  hexagon;  of  seven,  a 
heptagon ;  of  eight,  an  octagon  ;  of  nine,  a  nonagon ;  of  ten,  a 
decagon;  of  twelve,  n  dodecagon. 

258.  The  Perimeter  of  a  polygon  is  the  distance  around 
it,  or  the  sum  of  the  bounding  lines. 

259.  A  Salient  Angle  of  a  polygon  is  one  whose  sides, 
when  produced,  can  only  extend  without  the  polygon. 


POLYGONS, 


119 


260.  A  Re-entrant  Angle  of  a 

polygon  is  one  whose  sides,  when   pro- 
duced, can  extend  within  the  polygon. 

Illustration.— In  the  polygon  ABCDEFG, 
all  the  angles  are  salient  except  D,  which  is 
re-entrant. 

261.  A  Convex    Polygon    is  a 

polygon  which  has  only  salient  angles.  F«fl-  •«. 

A  polygon  is  always  supposed  to  be  convex,  unless  the  contrary 

is  stated. 

262.  A  Concave  or  Re-entrant  Polygon  is  a  polygon 

with  at  least  one  re-entrant  angle. 

263.  An  Equilateral  Polygon  is  a  polygon  whose  sides 
are  equal,  each  to  each ;  and  an  Equiangular  Polygon  is 
a  polygon  whose  angles  are  equal,  each  to  each. 


PROPOSITION    XV. 

264.  Theorem. — The  sum  of  the  interior  angles  of  a 
polygon  is  equal  to  twice  as  many  right  angles  as  the  poly- 
gon has  sides,  less  four  right  angles. 


Demonstration. 


Let  n  be  the  number  of  sides  of  any  polygon 

We  are  to  prove  that  the  sum  of  its  angles 
is  71  times  2  right  angles  less  4  right  angles. 

From  any  point  within,  as  0,  draw  lines  to 
the  vertices  of  the  angles.  As  many  triangles 
will  then  be  formed  as  the  polygon  has  sides, 
that  is,  n. 

The  sum  of  the  angles  of  the  triangles  is  n 
times  2  right  angles. 

But  this  sum  exceeds  the  sum  of  the  angles 
of  the  polygon  by  the  sum  of  the  angles  around 
the  common  vertex  0,  that  is,  by  4  right  angles. 


f-iy.  130. 


120 


ELEMENTARY    GEOMETRY. 


Hence  the  sum  of  the  angles  of  the  polygon  is 

7*  times  2  right  angles  less  4  right  angles,     q.  k.  d. 

265.  Scholium  1. — The  sum  of  the  angles  of  a  pentagon  is 

5  times  2  right  angles  -  4  right  angles,  or  6  right  angles. 

The  sum  of  the  angles  of  a  hexagon  is  8  right  angles ;  of  a  heptagon, 
10 ;  of  an  octagon,  12,  etc. 

266.  Scholium  2. — This  proposition  is  equally  applicable  to  triangles 
and  to  quadrilaterals.    Thus,  the  sum  of  the  angles  of  a  triangle  is 

3  times  2  right  angles  —  4  right  angles  =  2  right  angles. 
So  also  the  sum  of  the  angles  of  a  quadrilateral  is 

4  times  2  right  angles  —  4  right  angles,  or  4  riglit  angles. 

267.  Scholium  3. — To  find  the  value  of  an  angle  of  an  equiangular 
polygon,  divide  the  sum  of  all  the  angles  by  the  number  of  angles. 


PROPOSITION    XVI. 

268.  Theorem. — //  one  of  the  sides  of  a  polygon  is 
produced  (and  only  one)  at  each  vertex,  the  sum  of  the 
exterior  angles  thus  formed  is  four  right  angles. 

Demonstration. 

Let  n  be  the  number  of  the  sides  of  any 
polygon,  and  one  side  be  produced  at  each 
vertex. 

We  are  to  prove  that  the  sum  of  the  ex- 
terior angles  thus  formed,  &8  a  +  h  •\- c  -\-  d, 
etc.,  is  4  right  angles. 

At  each  of  the  n  vertices  there  are  two 
angles,  an  interior  and  an  exterior  one, 
whose  sum,  as  A  +  a,  is  2  right  angles. 
Hence  the  sum  of  all  the  exterior  and  inte- 
rior angles  is 

n  times  2  right  angl 


Fig.  131. 


REGULAR    POLYGONS, 


\%1 


Now,  from  this  sum  subtracting  the  sum  of  the  exterior  angles,  the 
remainder  is  the  sum  of  the  interior  angles. 

But,  by  the  preceding  propositiim,  4  right  angles  subtracted  from  n 
times  2  right  angles  leaves  the  sum  of  the  interior  angles. 

Therefore  the  sum  of  the  exterior  angles  is  4  right  angles,     q.  e.  d. 


OF    REGULAR     POLYGONS. 

269.  A  Regular  Polygon  is  a  polygon  which  is  both 
equilateral  and  equiangular  (263). 

270.  An  Inscribed  Polygon  is  a  polygon  whose  angles 
are  all  inscribed  in  the  same  circumference. 

271.  A  Circumscribed  Polygon  is  a  polygon  whose 
sides  are  all  tangent  to  the  same  circle.  The  circumference  is 
said  to  be  inscribed  in  the  polygon. 


PROPOSITION    XVII. 

272.  Theorem. — The  angles  of  an  inscribed  equilateral 
polygon  are  equal ;  and  the  polygon  is  regular. 

Demonstration. 

Let  ABCDEF  be  an  inscribed  poly- 
gon, having  AB  =r  BC  =  CD,  etc. 

We  are  to  prove  that  angle  ABC  = 
angle  BCD  =  angle  CDE,  etc. 

The  sides  of  the  polygon  being  equal 
chords,  subtend  equal  arcs  (151). 

Now  any  angle  of  the  polygon  is 
measured  by  ^  the  diflference  between 
the  circumference  and  the  sum  of  two 
of  these  equal  arcs,  as  angle  ABC  meas- 
ured by  ^  (circumference  —  arc  ABC) 
=  i  arc  AFEDC. 

Hence  all  the  angles  are  equal,  and  the  polygon  is  regular  (269).    Q.  E.  D. 
6 


Fig.  132. 


122 


ELEMENTARY    GEOMETRY. 


PROPOSITION    XVIII. 

273.   Theorem. — A  circumference    may   he    circum,- 
scribed  about  any  regular  polygon. 


Demonstration. 
Let  ABCDEF  be  a  regular  polygon. 

We  are  to  prove  that  a  circumference  can 
be  circumscribed  about  it. 

Bisecting  any  two  consecutive  sides,  as  FA 
and  AB,  by  perpendiculars,  as  Oa  and  06,  pass 
a  circumference  through  the  vertices  F,  A,  and 
B  (161). 

We  will  now  show  that  this  circumference 
passes  through  all  the  other  vertices. 

Revolve  the  quadrilateral  FO^A  upon  06  as 
an  axis  until  it  falls  in  the  plane  of  C06B,  6A  will  fall  in  its  equal  IB  (?) ; 
and  since  angle  A  =  angle  B,  and  side  AF  =  side  BO,  F  will  fall  in  C. 

Thus  it  appears  that  the  circumference  described  from  0,  and  pass- 
ing through  F,  A,  and  B,  also  passes  through  0. 

In  a  similar  manner  it  can  be  shown  that  the  same  circumference 
passes  through  all  the  vertices,  and  hence  is  circumscribed.     Q.  e.  d. 


PROPOSITION     XIX. 

274.  Theorem. — A  circumference  may  be  inscribed  in 
any  regular  polygon. 

Demonstration. 
Let  ABCDEF  be  a  regular  polygon. 

We  are  to  show  that  a  circumference  may 
be  inscribed  in  it. 

Let  0  be  the  centre  of  the  circumscribed 
circumference  (273);  then  the  sides  of  the 
polygon  are  equul  chords  of  this  circle,  and 
consequently  equally  distant  from  the  centre 
(160).  Fig  134. 


REOULAR    POLYGONS.  123 

Now  draw  the  perpendiculars  0«,  0&,  Oc,  Orf,  etc.  These  perpendic- 
ulars are  all  equal,  and  a  circumference  struck  from  0  as  a  centre,  with 
anyone  of  them,  as  Oa,  as  a  radius,  will  pass  through  5,  c,  d^  etc. 

Moreover,  the  sides  AB,  BC,  CD,  etc.,  being  perpendicular  to  the 
radii  Oa,  OJ,  etc.,  are  tangents  to  this  circumference,  which  is  therefore 
an  inscribed  circumference  (271).    Q.  e.  d. 

275.    Corollary. — The  centres  of  the  inscribed  and  cir- 
'  curnscrihed  circles  coincide. 


276.  The  Centre  of  a  regular  polygon  is  the  common  cen- 
tre of  its  inscribed  and  circumscribed  circles. 

277.  An  Angle  at  the  Centre  of  a  regular  polygon  is 
the  angle  included  by  two  lines  drawn  from  the  centre  to  the 
extremities  of  a  side,  as  FOA,  AOB  (Fig.  133). 

278.  The  Apotheni  of  a  regular  polygon  is  the  distance 
from  the  centre  to  any  side,  and  is  the  radius  of  the  inscribed 
circle. 


PROPOSITION    XX. 

279.  Theorem. — The  angles  at  the  centre  of  a  regular 
polygon  are  equal  each  to  each ;  and  any  one  is  equal  to 
four  right  angles  divided  by  the  number  of  sides  of  the 
polygon. 

Demonstration. 

Let  P  be  a  polygon  ofn  sides. 

We  are  to  prove,  Ist.  That  the  angles  at  the  centre  are  equal ;  and 

2d.  Tl.at  any  one  of  them  is  ^  ^ght  angles  _ 

n 

1st.  Each  angle  at  the  centre  intercepts  one  of  the  equal  sides  of  the 
polygon.  But  these  sides  are  chords  of  equal  arcs  (?).  Hence  the  sev- 
eral angles  at  the  centre  have  equal  measures,  and  are  therefore  equal. 

Q.  E.  D. 


124 


ELEMENTARY    GEOMETRY, 


2d.  The  sum  of  all  the  angles  at  the  centre  is  4  right  angles  (?),  and 
as  they  are  equal  and  n  in  number,  any  one  is 

4  right  angles 

2 5 Q.  E.  D. 


PROPOSITION    XXI. 

280.    Theorem. — Any  side  of  a  regular  inscribed  hex- 
agon is  equal  to  the  radius. 

Demonstration. 

Let  ABCDEF  be  a  regular  hexagon  inscribed  in  a  circle  whose  radius 
is  11. 

We  are  to  prove  that  any  one  of  the  equal 
sides,  as  AB,  equals  B. 

Let  0  be  the  centre  of  the  polygon,  and  draw 
OA,  OB,  etc. 

Now  in  the  triangle  AOB,  angle  0  is  ^  of  4 
right  angles,  or  ^  of  2  right  angles  (?). 

Whence  the  sum  of  the  angles  OAB  and  OBA 
is  I  of  2  right  angles  (?). 

But  the  triangle  AOB  is  isosceles,  OA  and 
OB  being  radii  of  the  same  circle.     Hence,  each  ^"'9-  '35. 

one  of  the  angles  at  the  base  is  ^  of  f  of  2  right  angles,  or  |  of  2  right 
angles.  Therefore  the  triangle  AOB  is  equiangular  and  consequently 
equilateral  (222),  and  AB  =  OA  =  i2.     Q.  e.  d. 


281.    A  Broken  Line  is  said  to  be  Convex  when  a  straight  line 
cannot  be  drawn  which  shall  cut  it  in  more  than  two  points. 


PROPOSITION    XXII. 

282.  Theorem. — A  convex  broken  line  is  less  than  any 
broken  line  which  envelops  it  and  has  the  same  extremities, 
the  former  lying  between  the  latter  and  a  straight  line 
joining  its  extremities. 


REGULAR    POLYGONS, 


125 


Demonstration. 

Let  ^hcd^  be  a  broken  line  enveloped  by  the  broken  line  ACDEFB, 
and  having  the  same  extremities  A  and  B. 

We  are  to  prove  that 

Mcd^  <  ACDEFB. 

Produce  the  parts  of  A5aZB  till  they 

meet  the  enveloping  line,  as  Kb  to  e,  he  to 
/,  and  cd  to  g. 

Now,  Kb-^he  <  ACe  (?), 

hc  +  cf  <he  +  eDE/(0, 

ed  +  dg  <cf  +ffg, 

dB  <  dg  -\-  gB. 


Fi{|.  136. 


Hence,  adding,  and  subtracting  common  terms, 

Ab  +  be  -^  cd  -{-  dB  <  ACe  +  cDEf  +  /Fg  +  gB, 
or  AbcdB  <  ACDEFB.     Q.  e.  d. 


angle. 


PROPOSITION    XXIII. 

Problem. — To  inscribe  a  circle  in  a  given  tri- 

SOLUTION. 


Let  ABC  be  a  triangle. 

We  are  to  inscribe  a  circle. 
Bisect  any  two  angles,  as  A  and  B  (208). 
From  the  intersection  of  the  bisectors,  as  0, 
let  fall  a  perpendicular,  as  OD* 

Then  is  0  the  centre  of  the  inscribed  circle, 
and  OD  its  radius. 

Hence  a  circle  described  with  0  as  a  centre 
and  OD  as  a  radius  will  be  inscribed. 


Fig.  137. 


Demonstration  of  Solution. 
From  0  let  fall  the  perpendiculars  OD,  OE,  and  OG  on  the  sides. 


126 


ELEMENTARY     GEOMETBY, 


Now  the  triangle  AOE  =  AOG  (?),  BEO  = 
BOD  (?). 

Hence  OD  =  OE  =  OG,  and  the  circum- 
ference struck  from  0  as  a  centre  with  a  radius 
OD,  passes  through  E  and  G. 

Moreover,  AC,  AS,  and  BC  are  perpendicu- 
lar to  the  radii  OG,  OE,  and  OD  respectively, 
and  hence  are  tangents  to  the  circle. 

Therefore  the  circle  is  inscribed  in  the 
triangle.    Q.  b.  d. 


Fig.  137. 


PROPOSITION    XXIV. 

284.  Problem. — In  a  given  circle  to  inscribe  a  square, 
and  hence  a  regular  octagon,  and  then  a  regular  polygon 
of  16  sides,  etc. 

[Let  the  pupil  give  the  solution  and  demonstration.] 


PROPOSITION    XXV. 

285.  Problem. — In  a  given  circle  to  inscribe  a  regular 
hexagon,  and  hence  an  equilateral  triangle  and  a  dodec- 
agon. 

[Let  the  pupil  give  the  solution.] 


PROPOSITION    XXVI. 

286.    Problem. — To    circumscribe   a  square  about  a 
given  circle. 

[Let  the  pupil  give  the  solution.] 


EXERCISES, 


127 


PROPOSITION    XXVII. 

287.    Problem. — To  circumscribe  an  equilateral  tri- 
angle about  a  circle. 

[Let  the  pupil  give  the  solution.] 


PROPOSITION    XXVIII. 

288.    Problem. — To  circumscribe  a  regular  hexagon 
about  a  given  circle. 

[Let  the  pupil  give  the  solution.] 


289.  Query. — Given  any  regular  inscribed  polygon,  how  is 
the  regular  cir(^m8cribed  polygon  of  the  same  number  of  sides 
constructed  ? 


EXERCISES 

290.  1.  Given  two  angles  of  a  tri- 
angle, to  find  the  third. 

Suggestions. — Tlie  student  should  draw 
two  angles  on  the  black lx)ard,  as  a  and  ft,  and 
then  proceed  to  find  the  third.  The  figure 
will  suggest  the  method.  The  third  angle 
is  c. 

The  solution   is  effected    also   by  con- 
structing the  two  given  angles  at  the  extrem-  Fig.  I38. 
ities  of  any  line,  and  producing  the  sides  till  they  meet  (?). 

2.  What  part  of  a  right  angle  is  one  of  the  angles  of  an  equi- 
lateral triangle  ?  From  this  fact,  how  can  you  obtain  an  angle 
equal  to  |  of  a  right  angle  ? 

3.  Two  angles  of  a  triangle  are  respectively  |  and  J  of  a  right 
angle.    What  is  the  third  angle  ? 


128  ELEMENTARY    GEOMETRY. 

4.  The  angles  of  a  triangle  are  respectively  f ,  J,  and  |  of  a 
right  angle.  Which  is  the  greatest  side?  Which  the  least? 
Can  you  tell  the  ratio  of  the  sides  ? 

5.  What  is  the  value  of  one  of  the  equal  angles  of  an  isosceles 
triangle  whose  third  angle  is  J  of  a  right  angle  ? 

6.  Two  consecutive  angles  of  a  quadrilateral  are  respectively 
I  and  f  of  a  right  angle,  and  the  other  two  angles  are  mutually 
equal  to  each  other.  What  is  the  form  of  the  quadrilateral  ? 
What  the  value  of  each  of  the  two  latter  angles? 

7.  One  of  the  angles  of  a  parallelogram  is  f  of  a  right  angle. 
What  are  the  values  of  the  other  angles  ? 

8.  The  two  opposite  angles  of  a  quadrilateral  are  respectively 
I  and  i  of  a  right  angle.  Can  a  circumference  be  circumscribed  ? 
If  so,  do  it. 

9.  Two  of  the  opposite  sides  of  a  quadrilateral  are  parallel, 
and  each  is  15  in  length.  What  is  the  figure  ?  Do  these  facts 
determine  the  angles  ? 

10.  Two  of  the  opposite  sides  of  a  quadrilateral  are  12  each, 
and  the  other  two  7  each.  What  do  these  facts  determine  with 
reference  to  the  form  of  the  figure  ? 

11.  What  is  the  value  of  an  angle  of  a  regular  dodecagon  ? 

12.  What  is  the  sum  of  the  angles  of  a  nonagon  ?  What  is 
the  value  of  one  angle  of  a  regular  nonagon  ?  Of  one  exterior 
angle  ? 

13.  What  is  the  regular  polygon,  one  of  whose  angles  is  l^nf 
right  angles  ? 

14.  What  is  the  regular  polygon,  one  of  whose  exterior  angles 
is  f  of  a  right  angle  ? 

15.  Can  you  cover  a  plane  surface  with  equilateral  triangles 
without  overlapping  them  or  leaving  vacant  spaces  ?  With 
quadrilaterals?  Of  what  form?  With  pentagons?  Why? 
With  hexagons?  Why?  What  insect  puts  the  latter  fact  to 
practical  use  ?  Can  you  cover  a  plane  surface  thus  with  regular 
polygons  of  more  than  6  sides  ?     Why  ? 


EXERCISES,  129 

THEOREMS    FOR    ORIGINAL.    INVESTI- 
GATION. 

[It  is  quite  desirable  that  students  have  exercise,  early  in  their  course, 
in  the  original  demonstration  of  theorems.  Those  which  are  given  in  this 
and  the  following  lists  are  not  such  as  are  essential  to  the  integrity  of  an 
elementary  course,  and  pupils  may  be  encouraged  to  demonstrate  more  or 
lesis  of  them,  as  their  time  and  ability  will  allow.  But  all  should  do  some 
such  work — it  is  the  true  test  of  mathematical  ability  and  attainment.] 

291.  1.  Theorem. — The  least  chord  that  can  he  drawn 
through  a  point  within  a  circle  is  the  chord  which  is  per- 
pendicular to  a  diameter  passing  through  the  same  point. 

2.  Theorem. — Tlie  shortest  distance  from  a  point 
without  a  circle  to  the  circumference  is  measured  in  a 
line  which  passes  through  the  centre. 

3.  Theorem. — The  sum  of  the  angles  form,ed  by  pro- 
ducing the  alternate  sides  of  any  pentagon  is  two  right 
angles. 

4.  Theorem. — Prove  that  the  sum^ 
of  the  angles  of  a  triangle  is  two  right 
angles,  by  producing  two  of  the  sides 
about  an  angle,  and  through  the  vertex 
of  this  angle  drawing  a  line  parallel  to 
the  third  side. 

Prove  the  same  by  producing  one 
side  of  the  triangle,  and  drawing  a  line 
through  the  vertex  of  the  exterior  angle 
parallel  to  the  non-adjacent  side. 

Fig.  139. 

5.  Theorem.—//  AB  is  any  chord,  AC  a  tangent  at  A, 
and  CDE  a  line  parallel  to  AB  and  cutting  the  circumfer- 
ence in  D  and  E,  the  triangles  ACD,  CAE,  and  ADB  are 
mutually  equiangular. 

6.  Theorem. — //  from  any  point  in  the  base  of  an 


130 


ELEMENTAR  Y     GEOMETR  Y. 


isosceles  triangle  lines  are  drawn  parallel  to  the  equal 
sides,  a  parallelogram  is  formed  whose  perimeter  is  equal 
to  the  sum  of  the  equal  sides. 


0UCTI0M  H% 


OF     EQUALITY. 

292.  Equality  signifies  likeness  in  every  respect. 

293.  The  equality  of  magnitudes  is  usually  shown  by  apply- 
ing one  to  the  other,  and  observing  that  the  two  coincide. 


OF     A  NGLES. 


PROPOSITION    I. 

294.  Theorem.— rM;o  angles  whose  corresponding 
sides  are  parallel,  and  extend  in  the  same  or  in  opposite 
directions  from  their  vertices,  are  equal. 

DEMONSTRATIOi?". 

Mrst,  In  («)  and  {a'),  let  B  and  E  be  two 
angles  having  BA  parallel  to  ED  and  extending  in 
the  same  direction  from  the  vertices,  and  also  BC 
parallel  to  EF  and  extending  in  the  same  direction 
from  the  vertices. 

We  are  to  prove  that  angles  B  and  E  are  equal. 

Produce  (if  necessary)  either  two  non-parallel 
sides,  as  BC  and  ED,  till  they  intersect,  as  in  H. 

ABC  r=  DHC  (?), 
and  DHC  =  DEF  (?). 

Therefore,  ABC  =  DEF.    Q.  e.  d. 


Fig.  140. 


JSQUALirr    OF   ANGLES, 


131 


Second.  In  (6)  and  (&'),  let  B'  and  E'  have 
B'A'  parallel  to  ET',  but  extending  in  an  oppo- 
site direction  from  the  vertices  ;  and  in  like  man- 
ner B'C  parallel  to^  but  extending  in  an  opposite 
direction  from  E'D'. 

We  are  to  prove  that  B'  and  E'  are  equal. 

Produce  (it*  necessary)  either  two  non -parallel 
sides,  as  A'B'  and  E'D',  till  they  meet  in  some 
point,  as  H'. 

D'H  B'  =  D  E  F'  (?), 
nnd  D'H'B'  =  A'B'C  (?). 

Therefore  D'ET'  =  A'B'C  (?).     H,  e.  d. 


Fig.  I4r. 


PROPOSITION    II. 

295.  Theorem.  —  Two  angles  having  their  corres- 
ponding sides  parallel,  ivhile  two  extend  in  the  same 
direction,  and  the  other  two  in  opposite  directions  from 
the  vertices,  are  supplemental. 

Demonstration. 


Let  ABC  and  DEF  be  two 

angles  whose  corresponding 
sides  BA  and  EF  are  parallel 
and  extend  in  the  same  direc- 
tion from  B  and  E,  while  BC 
and  ED  extend  in  opposite 
directions  from  the  vertices. 

We  are  to  prove  that  ABC 
and  DEF  are  supplemental. 

Produce  one  of  the  two  '^*   *  ' 

sides  having  opposite  directions  as  DE  to  H,  in  the  same  direction  from 
the  vertex  that  BC  extends. 

Now  DEF  is  supplemental  to  FEH  (?),  and  FEH  is  equal  to  ABC  (?) 

Therefore,  DEF  and  ABC  are  supplemental,    q.  e.  d. 


132  ELEMENTARY    GEOMETRY, 

^"  PROPOSITION    III. 

296.  Theorem. — If  the  sides  of  one  angle  are  perpen- 
dicular respectively  to  the  sides  of  another,  the  angles  are 
either  equal  or  supplemental. 

Demonstration. 

Let  ABC  be  any  angle  and  DE 
and  FH  be  two  lines  drawn  through 
any  point  0,  DE  being  perpendicular 
to  BC  and  FH  to  AB. 

We  are  to  prove  that  of  the  four 
angles  FOD,  DOH,  etc.,  two  are 
equal  to  ABC,  and  two  are  supple- 
mental. 

Draw  BS  bisecting  ABC,  and 
from  any  point  in  this  bisector,  as  L, 
draw  LM  and  LN,  respectively  paral- 
lel to  DE  and  FH. 

Now,  in  the  quadrilateral  LNBM,  the  sum  of  the  four  angles  is  four 
right  angles  (266) ;  and,  as  LNB  and  LMB  are  right  angles  (?),  NLM  and 
NBM  (or  ABC)  are  supplemental. 

But  NLM  =  FOD  (?)  =  HOE  C). 

Therefore  two  of  the  four  angles  FOD,  DOH,  etc.,  namely,  FOD  and 
HOE,  are  supplemental  to  ABC.     Q.  e.  d. 

Finally,  FOE  and  DOH  are  supplements  of  FOD  and  HOE  (?)  and 
hence  equal  to  ABC.     Q.  k.  d. 

297.  Scholium.— To  determine  whether  the  angles  are  equal,  or 
whether  they  are  supplemental,  we  may  consider  one  angle  as  moved 
(if  necessary)  till  its  vertex  falls  in  the  bisector,  its  sides  remaining 
paraMel  to  their  first  position.  Then,  if  both  sides  of  one  angle  extend 
towards,  or  both  extend  from  the  sides  of  the  other,  the  angles  are  sup- 
plemental, otherwise  they  are  equal. 


Fig.   143. 


OF    TRIANGLES. 


PROPOSITION    IV. 

298.  Theorem. — Two  triangles  which  have  two  sides 
and  the  included  angle  of  one  equal  to  two  sides  and  the 
included  angle  of  the  other,  each  to  each,  are  equal. 


EQUALITY    OF    TRIANGLES, 


133 


Demonstration". 


Let  ABC  and  DEF  be 
two  triangles,  having  AC 
=  DF,  AB  -  DE,  and  angle 
A  =  angle  D. 

We  are  to  prove  that 
the  triangles  are  equal. 

Place  the  triangle  ABC 
in  the  position  (6),  the  side 
AB  in  its  equal  DE,  and 
the  angle  A  adjacent  to  its 
equal  angle  D. 

Then  revolving  ABC  upon  DE,  until  it  falls  in  the  plane  on  the  oppo- 
site side  of  DE,  since  angle  A  =  angle  D,  AC  will  take  the  direction  DF; 
and  as  AC  =  DF,  C  will  fall  at  F.  Hence  BC  will  fall  in  EF,  and  the 
triangles  will  coincide.     Therefore  the  two  triangles  are  equal,     q.  e.  d. 


Scholium  1. — We  may  also  make  the  application  of  ABC  to 
DEF  directly.  The  method  here  given  is  used  for  the  purpose  of  uni- 
formity in  this  and  the  following.  We  may  observe  that  in  this,  as  in 
the  other  cases,  DB  is  perpendicular  to  FC,  and  bisects  it  at  0. 

300-  Scholium  2.  —This  proposition  signifies  that  the  two  triangles 
are  equal  in  all  respects^  L  «.,  that  the  two  remaining  sides  are  equal,  as 
CB  =  FE;  that  angle  C  =  angle  F,  angle  B  =  angle  E,  and  that  the 
areas  are  equal. 


PROPOSITION     V. 

301.  Theorem. — Tivo  triangles  which  have  two  angles 
and  the  included  side  of  the  one  equal  to  two  angles  and 
the  included  side  of  the  other,  each  to  each,  are  equal. 

Demonstration. 

Let  ABC  and  DEF  be  two  triangles,  having  angle  A  =  angle  D,  angle 
B  =  angle  E,  and  side  AB  =  side  DE. 


We  are  to  prove  that  the  triangles  are  equal. 


134 


ELEMENTARY    GEOMETRY. 


Place  ABC  in  the  po- 
sition (5),  the  side  AB  in 
its  equal  DE,  the  angle  A 
adjacent  to  its  equal  angle 
D,  and  B  adjacent  to  its 
equal  angle  E. 

Then  revolving  ABC 
upon  DE  till  it  falls  in  the 
plane  on  the  same  side  as 
DFE,    since     angle    A  =  Fig.  I45. 

angle  D,  AC  will  take  the 
direction  DF,  and  C  will  fall  somewhere  in  DF,  or  DF  produced. 

Also,  since  angle  B  =  angle  E,  BC  will  take  the  direction  EF,  and  C 
will  fall  somewhere  in  EF,  or  EF  produced. 

Hence,  as  C  falls  at  the  same  time  in  DF  and  EF,  it  falls  at  their  in- 
tersection F.  Therefore  the  two  triangles  coincide,  and  are  consequently 
equal.    Q.  e.  d. 

302.  Corollary. — //  one  triangle  has  a  side,  its  oppo- 
site angle,  and  one  adjacent  angle,  equal  to  the  correspond- 
ing parts  in  another  triangle,  the  triangles  are  equal. 

For  the  third  angles  are  equal  to  each  other,  since  each  is  the  supple- 
ment of  the  sum  of  the  given  angles.  Whence  the  case  is  included  in  the 
proposition. 


303.  Scholium. — A  triangle  may  have  a  side  and  one  adjacent  angle 
equal  to  a  side  and  an  adjacent  angle  in  another,  and  the  second  adjacent 
angle  of  the  first  equal  to  the  angle  opposite  the  equal  side  in  the  second, 
and  the  triangles  not  be  equal.     Thus,  in  the  figure,  AB  =  C'A',  A  =  A', 


and  B  =  B' ;  but  the  triangles  are  evidently  not  equal.     [Such  triangles 
are,  however,  simitar^  as  will  be  shown  hereafter.] 


EQUALITY    OF    TRIANGLES. 


135 


PROPOSITION    VI. 

304.  Theorem. — Two  triangles  which  have  two  sides 
and  an  angle  opposite  one  of  these  sides,  in  the  one,  equal 
to  the  corresponding  parts  in  the  other,  are  equal,  if  of 
these  two  sides  the  one  opposite  the  given  angle  is  equal  to 
or  greater  than  the  one  adjacent. 


Demonstration. 

In  the  triangles  ABC  and  DEF,  let  AC  =  DF,  CB  -  FE,  A  =  D,  and 

CB  (=  FE)>  AC  (=  DF). 
We  are  to  prove  that 
the  triangles  are  equal. 

Apply  the  triangle 
ABC  to  DEF,  placing  AC 
in  its  equal  DF,  the  point 
A  falling  at  D,  and  C  at 
F. 

Since  A  =  D,  AB 
will  take  the  direction 
DE. 

Let  fall  the  perpen- 
dicular FH  upon  DE,  or 
DE  produced. 

Now,  CB  being  ^  DF,  cannot  fall  between  it  and  the  perpendicular, 
but  must  fall  in  FD  or  beyond  both  (?). 

But  CB  cannot  fall  in  FD,  since  it  is  a  different  line  from  CA. 

Again,  as  CB  =  FE,  and  both  lie  on  the  same  side  of  FH,  they  must 
coincide  (114). 

Hence,    the   two   triangles   coincide,   and    are   consequently   equal. 

Q.  E.  D. 


PROPOSITION    VII. 


305.  Theorem. — Two  triangles  which  have  the  three 
sides  of  the  one  equal  to  the  three  sides  of  the  other,  each 
to  each,  are  equal. 


136 


ELEMENTART    GEOMETRY. 


Demon^stration. 


Fig.  148. 


l-et  ABC  and  DEF  be 
two  triangles,  in  which  AB 
=  DE,  AC  =  DF,  and  BC 
=  EF. 

We  are  to  prove  that 
the  triangles  are  equal. 

Place  the  triangle  ABC 
in  the  position  (5),  with 
the  longest  side,  AB,  in  its 
equal,  DE,  so  that  the 
other  equal  sides  shall  be 
adjacent,  as  AC  adjacent  to  DF,  and  BC  to  EF.  Draw  FC  cutting 
DE  in  0. 

Now,  since  AC  =  DF,  and  BC  =  EF,  DE  is  peipendicular  to  FC  at 
its  middle  point  (?). 

Hence,  revolving  ABC  upon  DE,  it  will  coincide  with  DEF  when 
brought  into  the  plane  of  the  latter,  since  OC  will  fall  in  OF  (?)  and  is 
equal  to  it. 

Therefore  the  two  triangles  coincide,  and  hence  are  equal,     q.  e.  d. 

306.  Corollary. — In  two  equal  triangles,  the  equal  an- 
gles lie  opposite  the  equal  sides. 


PROPOSITION    VIII. 

307.  Theorem.— //^m;o  triangles  have  two  sides  of  the 
one  respectively  equal  to  two  sides  of  the  other,  and  the 
included^  angles  unequal,  the  third  sides  are  unequal, 
and  the  greater  third  side  belongs  to  the  triangle  having 
the  greater  included  angle. 

Demonstration. 

Let  ABC  and  DEF  be  two  triangles  having  AC  =  DF,  CB  =  FE, 
and  C  >  F. 


We  are  to  prove  that  AB  >  DE. 


EQUALITY    OF    TRIANGLES. 


137 


Make  the  angle  ACE  =  DFE, 
take  CE  =  FE,  and  draw  AE. 
Then  is  the  triangle  ACE  =  DFE, 
and  AE  =  DE. 

Bisect  ECB  with  CH. 

Now  since  angle  DFE  =  ACE 
<  ACB  by  hypothesis,  CE  falls  be- 
tween CA  and  CB,  and  CH  will 
meet  AB  in  some  point,  as  H. 

Draw  HE. 

The  triangles  HCB  and  HCE 
have  two  sides  and  the  included 
angle  of  the  one,  equal  to  the  cor- 
responding parts  of  the  other, 
whence  HE  =  HB  (?). 

Now  AH  +  HE  >  AE 


Fig.  149. 


but 


Therefore, 


AH  +  HE  =  AH  +  HB  =  AB. 

AB  >  AE,  or  AB  >  DE.    q.  e.  d. 


308.  Corollary. —Conversely,  //  two  sides  of  one  tri- 
angle are  respectively  equal  to  two  sides  of  another,  and 
the  third  sides  are  unequal,  the  angle  opposite  this  third 
side  is  the  greater  in  the  triangle  which  has  the  greater 
third  side. 

That  is,  if  AC  =  DF,  CB  =  FE,  and  AB  >  DE,  angle  C  >  angle  F. 
For,  if  C  =  F,  the  triangles  would  be  equal,  and  AB  =  DE  (298) ;  and, 
if  C  were  less  than  F,  AB  would  be  less  than  DE,  by  the  proposition. 
But  both  these  conclusions  are  contrary  to  the  hypothesis.  Hence,  as  C 
cannot  be  equal  to  F,  or  less  than  F,  it  must  be  greater. 


PROPOSITION    IX. 

309.  Theorem.  —  Two  right-angled  triangles  which 
have  the  hypotenuse  and  one  side  of  the  one  equal  to  the 
hypotenuse  and  one  side  of  the  other,  each  to  each,  are 
equal. 


138 


ELEMENTARY     GEOMETRY, 


Demonstration. 


In  the  two  triangles  ABC  and  DEF,  right-angled  at  B  and  E,  let  AC 
DF,  and  BC  =  EF. 

We  are  to  prove  that  the 
triangles  are  equal. 

Place  FE  in  its  equal  CB, 
with  FD  on  the  same  side  of 
CB  that  AC  is. 

Then,  since  two  equal 
oblique  lines  cannot  be  drawn 
from  C  to  AB  on  the  same  side 
of  CB,  FD  will  coincide  with 
CA,  and  DE  with  AB  (?) 

Hence  the  two  triangles 
are  equal,  as  they  coincide 
throughout  when  applied  (292,  293).     Q.  e.  d. 


Fig.  150. 


PROPOSITION    X. 

310.  Theorem.  — Two  right-angled  triangles  having 
any  side  and  one  acute  angle  of  the  one  equal  to  the 
corresponding  parts  of  the  other  are  equal. 

Demonstration. 

One  acute  angle  in  one  triangle  being  equal  to  one  in  the  other,  the 
other  acute  angles  are  equal,  since  they  are  complements  of  the  same 
angles  (218).     The  case  then  falls  under  (301). 


EXERCISES. 

Exercise  1.  Given  the  sides  of  a  triangle,  as  15,  8,  and  5,  to 
construct  the  triangle. 

Ex.  2.  Given  two  sides  of  a  triangle,  a  =  20,  ^>  =  8,  and  the 
angle  B  opposite  the  side  h  equal  J  of  a  right  angle,  to  construct 
the  triangle. 


DETERMINATION    OF    TRIANGLES,  139 

Ex.  3.  Same  as  in  the  preceding  example,  except  h  =  12. 
Same,  except  that  b  =  25. 

Ex.  4.  Construct  a  triangle  with  angle  A  =  J  of  a  right 
angle,  angle  B  =  ^  of  a  right  angle,  and  side  a  opposite  angle 
A,  15. 

Ex.  5.  Construct  an  isosceles  triangle  whose  vertical  angle 
is  30^. 

Ex.  6.  Construct  a  right-angled  triangle  whose  hypotenuse 
is  12  and  one  of  whose  acute  angles  is  60^ 

Ex.  7.  Construct  an  equilateral  triangle,  and  let  fall  a  per- 
pendicular from  one  vertex  upon  the  opposite  side.  How  is  this 
angle  divided  ?  How  many  degrees  measure  the  angle  between 
the  perpendicular  and  one  side  ? 


THE  DETERMINATION  OF  POLYGONS. 

311.  A  triangle,  or  any  polygon,  is  said  to  be  Determined 
when  a  sufficient  number  of  parts  are  known  to  enable  us  to 
construct  the  figure,  or  to  find  the  unknown  parts.  If  two 
different  figures  can  be  constructed,  the  case  is  said  to  be 
Ambiguous. 

312.  Since,  in  such  a  case,  if  several  polygons  were  to  be  con- 
structed with  the  same  given  parts  all  would  be  equal,  the  condi- 
tions which  determine  a  polygon  are,  in  general,  the  same  as 
those  which  insure  equality  (292).  Hence,  having  shown  that 
certain  given  parts  determine  a  polygon,  we  may  assert  that  two 
polygons  having  these  parts  respectively  equal  are  equal,  except 
in  the  ambiguous  cases. 

PROPOSITION    XI. 

313.  Theorem. — A  triangle  is  determined  in  the  fol- 
lowing cases  : 

L   W7ien  two  sides  and  the  included  angle  are  known. 
II.   WTien  two  angles  and  the  included  side  are  known. 


140  ELEMENTARY    GEOMETRY, 

III.  When  the  three  sides  are  known. 

IV.  When  two  sides  and  an  angle  opposite  one  of  them 
are  known, 

(a.)  If  the  known  angle  is  right  or  obtuse. 

(&.)  If  the  known  angle  is  acute  and  the  known  side 
opposite  it  is  equal  to  the  perpendicular  upon  the  unknown 
side ;  or  equal  to  or  greater  than  the  other  known  side. 

{c.)  But,  if  the  known  angle  is  acute  and  the  known 
side  opposite  it  is  intermediate  in  length  between  the  other 
known  side  and  the  perpendicular  upon  the  unknown  side, 
the  case  is  ambiguous,  i.  e.,  there  are  two  triangles  possible. 

Demonstration. 

The  demonstration  of  this  proposition  is  effected  in  the  solution  of  the 
following  problems. 


314.   Problem.  —  Given  two  sides  and  the  included 
angle,  to  construct  a  triangle. 

Solution. 

Let  A  and  B  be  the  given  (or  known)  sides,  and  0  the  given  angle. 

We  are  to  construct  a  triangle  having 
an  angle  equal  to  0  included  between  sides 
equal  to  A  and  B. 

Draw  any  line,  as  O'D,  equal  to  either 
of  the  given  sides,  as  A. 

Lay  off  at  either  extremity  of  O'D,  as 
at  0',  an  angle  equal  to  0  (203),  and  make 
O'E  equal  to  B,  and  draw  ED. 

Then    will   EO'D   be  the  triangle  re-  Fig.  isi. 

quired. 

For,  if  two  triangles  (or  any  number)  be  constructed  with  the  sftme 
sides  and  included  angle,  they  will^all  be  equal  to  each  other  (298). 


315.  Problem. — Given  two  angles  and  the  included 
side,  to  construct  a  triangle. 


DETERMINATTNN    OF    TRIANGLES. 


141 


Solution. 


Let  M  and  N  be  the  two  given 
angles,  and  A  the  given  side. 

We  are  to  construct  a  triangle 
having  a  side  equal  to  A  and  in- 
cluded between  the  vertices  of  two 
angles  equal  respectively  to  M 
and  N. 

Draw  DE  equal  to  A.  At  one 
extremity,  as  D,  make  angle  FDE  = 
M,  and  at  E  make  FED  =  N. 

Then  is  DEF  the  triangle  re- 
quired (?). 


QuKRY. — What  is  the  limit  of  the  sum  of  the  given 


316.  Problem. 

angle. 


Given  three  sides,  to  construct  a  tri- 


Solution. 


Let  A,  B,  and  C  be  the  three  given  sides. 

We  are  to  construct  a  triangle  which 
shall  have  its  three  sides  respectively 
equal  to  A,  B,  and  C. 

Draw  DE  =  A. 

With  D  as  a  centre  and  a  radius 
equal  to  B,  strike  an  arc  intersecting  an 
arc  struck  from  E  as  a  centre,  with  a 
radius  C. 

The  triangle  DEF  is  the  triangle 
sought  (?). 


Fig.   153. 


317,  ScHOLiDM.— If  any  one  of  the  three  proposed  sides  is  greater 
than  the  sum  or  less  than  the  difference  of  the  other  two,  a  triangle  is 
impossible  (?). 


318.  Problem. — To  construct  a  triangle,  having  given 
two  sides  and  the  angle  opposite  one  of  them. 


142 


ELEMENTARr    GEOMETBr. 


Solution. 


There  are  three  cases. 


CASE  (a). 
When  the  given  angle  is  right  or  obtuse. 

Let  0  be  the  angle,  and  A  and  B 
the  sides,  the  angle  0  to  be  opposite 
the  side  A. 

Construct  angle  NDM  =:  0  (203), 
and  take  FD  =  B. 

From  F  as  a  centre,  with  A  as  a 
radius,  strike  an  arc  cutting  DM  in 
E,  and  draw  FE. 

Then  is  FDE  the  triangle  sought. 

For  it  has  FD  =  B,  FE  =  A 
(since  FE  is  a  radius  of  a  circle 
struck  with  A  as  a  radius),  and  angle 
FDE,  opposite  FE,  equal  to  0. 

If  the  given  angle  were  right,  the  construction  would  be  the  same. 


Fig.  154. 


CASE  (h). 

When  the  given  angle  is  acute,  and,  1st,  the  side  opposite 
equal  to  the  jterpendicular  upon  the  unknoivn  side,  and,  2d, 
when  the  side  opposite  is  equal  to  or  greater  than  the  other 
given  side, 

1st,  Let  A  and  B  be  the  given  sides 
and  0  the  given  angle  opposite  B. 

Proceed  exactly  as  in  the  preceding 
case,  but  when  the  arc  is  struck  from  F 
as  a  centre  with  a  radius  equal  to  B, 
instead  of  intersecting  DM  it  will  be 
tangent  to  it,  since  B  =  FE  is  the  per- 
pendicular, and  a  line  which  is  perpen- 
dicular to  a  radius  at  its  extremity  is 
tangent  to  the  arc  (156). 


2d,  If  the  side  opposite  the  given  angle  is  equal  to  the  other 
given  side,  the  arc  struck  from  F  tvith  it  as  a  radius  will  cut 


DETERMINATION    OF    QUADRILATERALS. 


143 


DM  at  an  equal  distance  tuith  FD  from  the  foot  of  the  perpen- 
dicular (?),  and  tite  trUmgle  formed  will  he  isosceles  {?), 

If  the  side  opposite  is  greater  than  the  other  given  side,  it 
tu^iU  cut  MD  hut  once  (?)  and  there  ivUl  he  hut  one  triangle. 


CASE   (r). 

When  the  given  angle  is  acute,  and  the  given  side  opposite 
it  is  intermediate  in  length  hetween  the  other  given  side  and 
the  perpendicular  to  the  unknown  siile. 


Let  A  and  B  be  the  given  sides 
and  0  the  angle  opposite  B,  B 
being  intermediate  in  length  be- 
tween A  and  the  perpendicular 
FH  on  the  unknown  side. 

Proceed  as  in  the  two  preced- 
ing cases,  but  instead  of  tangency 
we  get  two  intersections  of  DM 
by  the  arc  struck  from  F  with 
radius  B.  as  E  and  E',  since  two 
equal  oblique  lines  can  be  drawn 

from  F  to  DM  (114),  and  B  being  p;g   ,55^ 

less   than   FD  =  A,   FE    will   lie 
between  FD  and  FH,  and  FE'  beyond  FH  (113). 

Thus  we  have  two  triangles,  DEF  and  DE'F,  each  of  which  fulfills  the 
required  conditions. 


319.  Scholium. — In  order  that  the  triangle  should  be  possible,  the 
side  opposite  the  given  angle  must  be  equal  to  or  greater  than  the  per- 
pendicular upon  the  unknown  side. 


OF    QUADRILATERALS. 

The  subject  of  the  conditions  which  determine  a  quadrilateral  or 
other  polygon  is  quite  an  important  and  practical  subject,  especially  in 
surveying,  and  we  treat  the  problem  of  the  equality  of  polygons  of  more 
than  three  sides  in  this  way.     (See  312.) 


144 


ELEMENTARY    GEOMETRY, 


PROPOSITION    XII. 

320.  Theorem. — J_  quadrilateral  is  determined  when 
there  are  given  in  their  order  : 

I.  The  four  sides  and  either  diagonal, 

II.  The  four  sides  and  one  angle. 

III.  1st.   Three  sides  and  two  included  angles. 

2d.  When  the  two  angles  are  not  both  included  between 
the  known  sides,  the  case  may  be  ambiguous. 

IV.  Three  angles  and  two  sides,  the  unknown  sides 
being  non-parallel. 

Demonstration. 

CASE    I. 

Let  a,  hf  c,  d  (Fig.  157),  be  the  sides  in  order,  and  e  the  diagonal 
joining  the  vertex  of  the  angle  between  a  and  d  with  the  vertex  between 
b  and  c. 

With  LO  =  a,  OM  =  J,  MN  =  e,  NL  =  d^  and  LM  =  e,  construct,  by 
(316),  the  triangles  LOM  and  MNL,  on  LM  as  a  common  side. 
Then  is  LOMN  the  quadrilateral  sought. 


Fig.  157. 


Fig.  158. 


CASE     II. 

Let  <i,  b,  c,  and  d  (Fig.  158),  be  the  given  sides  in  order,  and  0  the 
angle  included  between  a  and  b. 

With  the  same  notation  as  before,  construct  the  triangle  LOM  by 
(314),  and  then  LMN  by  (316),  and  the  quadrilateral  is  constructed,  i.e., 
all  the  parts  are  found. 


DETERMINATION    OF    QUADRILATERALS. 


146 


CASE    III. 


Let  a,  b,  and  c  (Fig.  159) 
be  the  given  sides  in  order. 

1st,  Let  both  the  given 
angles  0  and  M  be  in- 
eluded  between  the  given 
sides,  0  being  included  by 
by  a  and  b,  and  !A  by  b 
and  c. 

Construct  an  angle  LOM 
=  0,  and  take  OL  =  o  and 
OM  =  5. 

Now  lay  off  the  angle  OMN 
=  M,  and  taking  MN  =  c, 
draw  LN. 

Then  is  LOMN  the  quadri- 
lateral Bought. 


Fig.  I5». 


2d,  Ambiguous  Cases.  —  If  three  sides  and  two  angles  of 
a  quadrilateral  are  given,  and  both  the  given  angles  are  not 
included  between  given  sides,  the  case  may  be  Ambiguous. 

There  may  be  three  cases :  1st.  "When  the  two  given  angles  are  con- 
secutive, and  one  only  is  included  between  given  sides;    2d.  When  the 
given  angles  are  consecutive,  and  the  includ- 
ed side  is  unknown ;   3d.  When  the  given 
angles  are  opposite. 

Fig.  160  shows  how  an  ambiguous  solu- 
tion may  arise  under  Case  1.  The  given 
parts  are  a,  &,  c,  and  angles  L  and  0. 

Fig.  161  shows  how  such  solutions 
may  arise  under  cases  2  and  3. 


Fig.  161. 


146  ELEMENTARY    GEOMETRY, 

CASE     IV. 

IsU  Let  the  three  given  angles  be  0,  M,  atid  N,  and,ftr8tf 
let  u  and  h  he  two  consecutive  given  sides. 

Since  the  sum  of  the  angles  of  a  quadrilateral  is  4  right  angles,  and 
0,  M,  and  N  are  given,  the  fourth,  L,  can  be  found  (241). 
[Let  the  student  make  the  construction.] 

2d,  Let  0,  M,  avid  N  he  the  given  angles,  and  a  and  c  the 
given  non-consecutive  sides,  d  and  h  helug  non-parallel, 
i.  e,,  the  angles  L  and  0  iiot  helng  supplemental. 

Find  the  fourth  angle  by  subtract- 
ing the  sum  of  the  three  given  angles 
from  4  right  angles.  Whence  all  the 
angles  are  known. 

Lay  off  side  a  and  at  its  extremities 
make  LOX  =  0,  and  OLY  =  L. 

Then  draw  any  line,  Arw,  making 
the  angle  m  =  M. 

Take  ink  =  c,  and  through  A  draw 
AS  parallel  to  OX.     Let  this  intersect 
LY  in  N.     Through  N  draw  NM  parallel    ^^^^^    ^.     ^^^ 
to  Am. 

Then  is  NM  =  the  given  side  c  (i),  and  OMN  =  the  given  angle  M  (?)^ 
and  LNM  =  the  given  angle  N  (?). 

Hence  LOMN  is  the  required  quadrilateral. 

321.  Scholium.— With  a  given  set  of  parts,  as  above,  the  possibility 
of  constructing  a  quadrilateral  can  be  determined  on  the  same  principle 
as  the  possibility  of  a  triangle. 

1.  In  Case  I,  if  the  diagonal  is  less  than  the  sum  and  greater  than  the 
difference  of  the  sides  of  either  of  the  two  triangles  into  which  it  divides 
the  quadrilateral,  the  quadrilateral  is  possible,  but  not  otherwise. 

2.  In  Case  II,  the  two  given  sides  and  their  included  angle  always 
make  a  triangle  possible;  whence  the  possibility  of  the  quadrilateral  will 
be  determined  by  the  relation  of  the  other  two  sides  to  the  third  side  of 
this  triangle,  as  (Fig.  158)  when  c -^  d  >  LM,  and  c-d  <  LM,  the 
quadrilateral  is  possible,  but  not  otherwise. 

3.  In  Case  III,  the  Ist  problem  is  always  possible.  The  student  will 
be  able  to  determine  when  the  several  cases  in  the  2d  are  possible  by  in- 
specting Figs.  160  and  161. 

4.  In  Case  I\',  the  first  problem  is  always  possible  when  the  sum  of 


^?°f7^ 

^f] 

K 

-^0 

U 

-■-'"'^CA^ 

Kv    ^^ 

DETERMINATION    OF   POLYGONS.  147 

the  given  angles  is  less  than  4  right  angles.  In  the  second  problem,  if 
the  unknown  sides  are  parallel,  the  problem  is  indeterminate,  i.  e.,  there 
may  be  any  number  of  solutions,  if  any. 

Note. — In  problems  of  this  class,  it  is  usually  understood  that  the 
given  parts  are  such  as  to  allow  the  construction ;  i.  c.,  that  they  are  parts 
of  a  possible  polygon. 

322.  Corollary  1. — A  parallelogrartv  is  determined 
when  two  sides  and  their  included  angle  are  given. 

Since  the  opposite  sides  of  a  parallelogram  are  equal  (250),  all  the 
sides  are  known  when  two  are  given,  and  the  case  falls  under  Case  II  of 
the  proposition, 

323.  Corollary  2. — Ihvo  rectangles  having  equal  bases 
and  equal  altitudes  are  equal. 

Exercise  1.  Construct  a  quadrilateral  three  of  whose  con- 
secutive sides  are  20, 12,  and  15,  and  the  angle  included  between 
20  and  the  unknown  side  f  of  a  right  angle,  and  that  between 
15  and  the  unknown  side  J  a  right  angle. 

Ex.  2.  Construct  a  quadrilateral  three  of  whose  sides  shall  be 
5,  4.2,  and  4,  and  in  which  the  angle  between  the  unknown  side 
and  the  side  5  shall  be  J^  of  a  right  angle,  and  that  between  the 
unknown  side  and  side  4,  1 J  right  angles.  How  many  solutions 
are  there?  How  many  solutions  if  the  second  side  is  made  1.2, 
and  the  third  2  ?  How  many  if  the  second  side  is  made  1,  and 
the  third  1.5  ? 


OF    PO  L YGO  NS 


PROPOSITION   XIII. 

324.  Theorem. — A  polygon  is  determined  when  two 
consecutive  sides,  the  diagonals  from  the  vertex  of  their 
included  angle,  and  the  consecutive  angles  included  he- 
tween  these  lines  are  given. 

[Let  the  student  show  how  the  construction  is  made,  and  thus  demon- 
strate the  proposition.] 


148  ELEMENTARY    GEOMETRY, 


PROPOSITION    XIV. 

325.    Theorem.—./^  polygon  is  determined  by  vieans 

of  its  sides  and  angles,  when  there  are  given  in  order  : 
I.  All  the  parts  except  two  angles  and  their  included  side. 
TI.  All  the  parts  except  three  angles. 
III.  All  the  parts  except  two  non-parallel  sides. 

C  o  If  s  T  R  u  c  T  I  0  N  s . 

CASE    I. 

Beginning  at  one  extremity  of  the  unknown  side,  and  constructing 
the  given  sides  and  angles  in  order  till  all  are  constructed,  and  joining 
the  extremities  of  the  broken  line  thus  drawn,  the  polygon  will  be  con- 
structed. 

CASE    II. 
Ist,  When  the  three  angles  lire  consecutive. 

Suppose  the  polygon  to  be  ABCDEFG,  and  the  unknown  angles  A,  G, 
and  F.  Commencing  with  side  AB,  lay  oflf  the  given  sides  and  angles  in 
order  till  the  unknown  angle  F  is  reached.  Then  from  F  as  a  centre, 
with  a  radius  equal  to  the  known  side  FG,  strike  an  arc  intersecting  an 
arc  struck  from  A  as  a  centre  with  the  side  AG  as  a  radius.  This  inter- 
section determines  the  remaining  vertex  of  the  polygon. 

Query  — When  does  this  case  become  impossible  ? 

2d,  When  two  of  the  mi  known  angles  are  consecutive  and 
the  third  is  separated  from  both  the  others. 

Let  A,  B,  and  F  be  the  unknown  angles. 
The  two  partial  polygons  AiHGF  and 
BCDEF  can  be  constructed,  and  thus  the 
sides  AF  and  BF  will  become  known,  as 
also  the  angles  AFG,  lAF,  BFE,  and 
FBC. 

Then  constructing  the  triangle  ABF, 
whose  three  sides  are  now  known,  the 
angles  AFB,  ABF,  and  FAB  become  known. 
Hence  all  the  parts  of  the  polygon  are 
found,  for  ^'9-  '"• 

the  angle  GFE  =  AFG  +  AFB  -f-  BFE,  etc. 


DETERMINATION    OF    POLYGONS. 


149 


3d,  When  no  two  of  the  three  unknown  angles  are  con- 
secutive. 


Let  A,  C,  and  F  be  the   unknown 


Constructing  the  broken  lines  ABC, 
CDEF,  and  FGHIA  separately,  and 
apart  from  the  position  where  the 
polygon  is  to  be  constructed,  the  diago- 
nals which  form  the  sides  of  the  triangle 
ACF  can  be  determined  by  joining  the 
extremities  A  and  C,  C  and  F,  and  F 
and  A. 

This  triangle  can  then  be  con- 
structed in  the  position  desired,  and 


Fig.  164. 


the  broken  lines  constructed  on  its  sides,  as  in  the  figure 

CASE    III. 

Under  this  case  we  have  two  problems: 

1st,    When  the  tivo  unknotvn  sides  are  consecutive, 

2d,    When  the  two  unknown  sides  are  separated. 


[The  student  will  be  able  to  effect  the  construction.  The  first  is 
similar  to  that  of  Case  II,  1st  problem.  The  second  is  effected  by 
obtaining  a  quadrilateral  similarly  to  the 
construction  in  Case  II,  3d  problem. 

326-  In  case  the  unknown  parts  are 
two  parallel  sides,  as  a  and  6,  it  is  evident 
that  these  may  be  varied  in  length  at  plea- 
sure without  changing  the  value  of  the  other 
parts.  ' 

327-  It  will  be  a  profitable  exercise  for 
the  student  to  reduce  the  determination  of  polygons  to  that  of  quadri- 
laterals, and  both  to  that  of  triangles. 


150  ELEMENTARY     GEOMETRY, 

PROPOSITIONS    FOR    ORIGINAL    SOLU- 
TION    AND    DEMONSTRATION. 

328.  1.  Theorem,— IT^e  sum  of  the  exterior  angles  of 
a  polygon  is  four  right  angles. 

Prove  by  drawing  lines  from  a  point  and  parallel  to  the  sides  of  the 
polygon. 

2.  Theorem. — The  sum  of  the  angles  of  a  polygon  is 
twice  as  many  right  angles  as  the  polygon  has  sides, 
less  four  right  angles. 

Having  proved  the  jweceding,  base  the  proof  of  this  upon  that. 

3.  Theorem. — //  the  sum  of  two  opposite  sides  of  a 
quadrilateral  is  equal  to  the  sum  of  the  other  two  op- 
posite sides,  show  that  a  circle  can  be  inscribed  in  the 
quadrilateral. 

4.  Theorem. — //  from,  a 
point  without  a  circle  two 
tangents  are  drawn,  and  also 
a  chord  joining  the  points  of 
tangency,  the  angle  included 
between  a  radius  drawn  to 
either  point  of  tangency  and 
the  chord  is  half  the  angle 
included  between  the  tan- 
gents. ^''^'  "'• 

5.  Theorem. — In  an  isosceles  triangle  the  line  drawn 
from  the  vertex  to  the  middle  of  the  base  bisects  the 
triangle  and  also  the  angle  at  the  vertex. 

6.  Problem. —  With  a  given  radius  draw  a  circle  tan- 
gent to  the  sides  of  a  given  angle. 

7.  Problem. — Through  a  given  point  within  a  given 
angle  draw  a  line  which  shall  make  equal  angles  with 
the  sides. 


EQUIVALENCY   AND    AREA. 


151 


8.  Problem. — To  draw  a  circumference  through  two 
given  points  and  having  its  centre  in  a  given  line;  or, 
to  find  in  a  given  line  a  point  equally  distant  from  two 
points  out  of  that  line. 


•9.  Theorem.—//  from  the  ^ 
extremities  of  a  diameter  per- 
pendiculars are  let  fall  ot^  any 
secant,  the  parts  intercepted 
between  the  feet  of  these  per- 
pendiculars and  the  circum- 
ference are  equal. 


Fig.  167. 


10.   Problem. — To  trisect  a  right  angle. 

Suggestion. — What  is  the  value  of  an  angle  of  an  equilateral  tri- 
angle ? 


OF     EQUIVALENCY     AND     AREA. 

329.  Equivalent  Figures  are  such  as  are  equal  in  mag- 
nitude. 

330.  The  Area  of  a  surface  is  the  number  of  times  it  con- 
tains some  other  surface  taken  as  a  unit  of  measure ;  or  it  is  the 
ratio  of  one  surface  to  another  assumed  as  a  standard  of  measure. 


PROPOSITION    I. 

331.    Theorem. — Parallelograms  having  equal  bases 
and  equal  altitudes  are  equivalent. 


152  elementary  geometry. 

Demonstration. 

Let  ABCD  and  EFGH  be  two  parallelograms  having  equal  bases,  BC 
and  FG,  and  equal  altitudes. 

We  are  to  prove 
that  the  parallelo- 
grams are  equivalent. 

Apply  EFGH  to 
ABCD,  placing  FG  in 
its    equal    BC;     and, 

since  the  altitudes  are  ^'9' 

equal,  the  upper  base  EH  will  fall  in  AD  or  AD  produced,  as  E'H'. 

Now,  the  two  triangles  AE'B  and  DH'C  are  equal,  since  they  have  two 
sides  and  the  included  angle  of  the  one  equal  to  two  sides  and  the  in- 
cluded angle  of  the  other;  viz.,  AB  =  DC,  being  opposite  sides  of  a 
parallelogram;  and  for  a  like  reason  BE'  =  CH'.  Also,  angle  ABE'  = 
angle  DCH',  by  reason  of  the  parallelism  of  their  sides  (294)- 

These  triangles  being  equal, 

the  quadrilateral  ABCH'  —  the  triangle  AE'B  =  ABCH'  —  DH'C. 
But  ABCH'  -  AE'B  =  E'BCH'  =  EFGH; 

and  ABCH'  -  DH'C  =  ABCD. 

Hence,  ABCD  =  EFGH.     Q.  e.  d. 

332.  Corollary. — Any  parallelogram  is  equivalent  to  a 
rectangle  having  the  same  base  and  altitude. 


PROPOSITION    II. 

333.  Theorem. — A  triangle  is  equivalent  to  one-half 
of  any  parallelogram  having  an  equal  base  and  an 
equal  altitude  with  the  triangle. 

Demonstration. 

Let  ABC  (Fig.  169)  be  a  triangle. 

We  are  to  prove  that  ABC  is  equivalent  to  one-half  a  parallelogram 
having  an  equal  base  and  an  equal  altitude  with  the  triangle. 


EQUIVALENCY    AXD    AREA 


153 


Consider  AB  as  the  base  of  the  triangle, 
and  complete  the  parallelogram  ABCD  by 
drawing  AD  parallel  to  EC,  and  DC  to  AB. 

Now  ABCD  has  the  same  base,  AB,  as 
the  triangle,  and  the  same  altitude,  since 
the  altitude  of  each  is  the  perpendicular 
distance  between  the  parallels  DC  and  AB. 

But  ABC  is  half  of  ABCD  (249),  and  as  ^'9-  '^^• 

any  other  parallelogram  having  an  equal  base  and  altitude  with  ABCD  is 
equivalent  to  ABCD  (331),  ABC  is  equivalent  to  one-half  of  any  parallel- 
ogram having  an  equal  base  and  altitude  with  ABC.    Q.  e.  d. 

334.  Corollary  1. — J.  tHangle  is  equivalent  to  one-half 
of  a  rectangle  having  an  equal  base  and  an  equal  al- 
titude with  the  triangle. 

335.  Corollary  2.~  Triangles  of  equal  bases  and  equal 
altitudes  are  equivalent,  for  they  are  halves  of  equivalent 
parallelograms. 


PROPOSITION  III  . 


336.  Theorem. — The  square  described  on  n  times  a 
line  is  n^  times  the  square  described  on  the  line,  n  being 
any  integer. 

Demonstration. 

Let  u  be  any  line  and  AB  a  line  n  times  as  long,  n  being  any  integer. 

We  are  to  prove  that  the  square  de- 
scribed on  AB  is  V?  times  the  square 
on  Aa. 

Construct  on  AB  the  square  ABCD. 

Since  m  is  a  measure  (76)  of  AB,  by 
hypothesis,  divide  AB  into  n  equal  parts 
by  applying  m,  and  at  the  points  of  di- 
vision a,  ft,  c,  etc.,  draw  parallels  to  AD. 

In  like  manner  divide  AD,  and  draw 
through  the  points  of  division  a',  5',  c', 
etc.,  parallels  to  AB. 

Then  are  the  surfaces  1,  2,  3,  4,  5,  6,  Pi^.  170 


154 


ELEMENTAR  Y    GEOMETB T, 


etc.,  squares,  since  their  opposite  sides 
are  parallel  (139)  and  equal  (138),  and 
their  angles  are  right  angles  (125)^ 

Now  of  these  squares  there  are  n  in 
each  of  the  rectangles  a'B,  5'E,  etc.  (?), 
and  as  there  are  n  divisions  in  AD,  there 
are  n  rectangles. 

Hence  there  are  n  times  n,  or  n^ 
squares  in  ABCD.     q.  e.  d. 

Fig.  170. 

337.  Corollary. — The  square  described  on  twice  a  line 
is  four  times  the  square  described  on  the  line ;  that  on  3 
times  a  line  is  9  times  the  square  on  the  line,  etc. 


PROPOSITION    IV. 

338.  Theorem. — A  trapezoid  is  equivalent  to  two  tri- 
angles having  for  their  bases  the  upper  and'  lower  bases  of 
the  trapezoid,  and  for  their  common  altitude  the  altitude 
of  the  trapezoid. 

By  constructing  any  trapezoid,  and  drawing  either  diagonal,  the  stu- 
dent can  show  the  truth  of  this  theorem. 


PROPOSITION    V. 


339.   Problem.- 

lent  triangle. 


To  reduce  any  polygon  to  an  equiva- 


SOLUTION. 


Let  ABCDEF  (Fig.  171)  be  a  polygon. 

We  are  to  reduce  it  to  an  equivalent  triangle. 

Draw  any  diagonal,  as  EC,  between  two  alternate  vertices,  and  through 
the  intermediate  vertex,  D,  draw  DH  parallel  to  EC  and  meeting  BC  pro- 
duced in  H.    Then  draw  EH. 


EQUIVALENCY   AND    AREA, 


155 


In  like  manner,  draw 
FH,  and  through  E  draw  El 
parallel  thereto,  meeting 
BH  produced  in  I.  Then 
draw  Fl. 

Again,  draw  the  diag- 
onal FB,  and  through  A 
draw  AG  parallel  thereto, 
meeting  BC  produced  in  G. 
Then  draw  FG. 

Now  FGI  is  equivalent  to  ABCDEF 


N 


K  ^M\ 


Fig.  171. 


Demonstration  of  Solution. 

Consider  the  polygon  ABCDEF  as  diminished  by  ECD  and  then  in- 
creased by  ECH.  Since  these  triangles  have  the  same  base  EC,  and  the 
same  altitude  (as  their  vertices  lie  in  DH  parallel  to  EC,  and  parallels  are 
everywhere  equidistant),  the  triangles  are  equivalent  (335)-  Hence, 
ABHEF  is  equivalent  to  ABCDEF  (?). 

In  like  manner  ABIF  is  equivalent  to  ABHEF,  and  FGI  to  ABIF. 

Hence  FGI  is  equivalent  to  ABCDEF.    q.  e.  d. 


A  R  E  A. 


340.  An  Infinitesimal  is  a  quantity  conceived  under  such 
a  law  as  to  be  less  than  any  assignable  quantity. 

Illustration.— Consider  a  line  of  any  finite  length,  as  one  foot. 
Conceive  this  line  bisected,  and  one-half  taken.  Again  conceive  this  half 
bisected,  and  one-half  of  it  taken.  By  this  process  it  is  evident  that  the 
line  may  be  reduced  to  a  line  less  than  any  assignable  line.  Moreover,  if 
the  process  be  considered  as  repeated  infinitely,  the  result  is  an  infini- 
tesimal. 

Tins  is  the  familiar  conception  of  the  last  term  of  a  decreasing  infinite 
progression,  the  last  term  of  which  is  called  zero. 


341.    Principle  I. — In  comparison  with  finite  quanti- 
ties, an  infinitesimal  is  zero. 


156  ELEMENTARY     GEOMETRY, 

Thus,  suppose  —  =  a, 

w,  n,  and  a  being  finite  quantities.    Let  i  represent  an  infinitesimal ;  then 

m  ±  i  m  m  ±  i 

,     or    .,    or    :, 

n     '  n  ±1  n±i 

is  to  be  considered  as  still  equal  to  a,  for  to  consider  it  to  difi*er  from  a 

by  any  amount  we  might  name,  would  be  to  assign  some  value  to  *'. 

342.  Principle  II. — Any  two  geometrical  magnitudes 
of  the  sam^e  kind  are  to  he  conceived  as  commensurable  hy 
an  infinitesimal  unit. 

By  the  process  for  obtaimng  the  common  measure  of  two  lines  (84), 
the  remainder  may  be  made  (in  conception)  less  than  any  assignable  quan- 
tity, and  hence  in  comparison  with  the  lines  should  be  considered  zero. 

The  same  conception  may  be  applied  to  any  geometrical  magnitudes. 


PROPOSITION    VI. 

343.    Theorem.-  Rectangles  are  to  each  other*  as  the 
products  of  their  respective  bases  and  altitudes. 

First  Demonstration. 

Lemma. — Two  rectangles  of  equal  altitudes  are  to  each 
other  as  their  bases. 

Let  ABCD  and  abed  be  two  rectangles  having  their  altitudes  AD  and 
ad  equal. 

Suppose  rectangle  ABCD  gen- 
erated by  the  movement  of  AD  from 
AD  to  BC,it  remaining  all  the  time 
parallel  to  its  first  position,  and 
suppose  ahcd  generated  in  like  man- 
ner by  the  movement  of  ad. 

Let  these  equal  generatrices  AD  ^'3'  '^^• 

and  ad  move  with  uniform  and  equal  velocities ;  then  it  is  evident  that 
the  surfaces  generated  will  be  as  the  distances  AB  and  ab. 

That  is,  ABCp^AB. 

abed         ab 

*  This  is  a  common  elliptical  form,  meaning  that  surfaces,  or  areas,  are 
to  each  other. 


EQUIVALENCY  AND    AREA, 


167 


Now  let  M  and  N  be  any  two  rectan- 
gles, the  base  of  M  being  AB  and  the 
altitude  BC,  and  the  base  of  N  BE  and 
its  altitude  BG. 

We  are  to  prove  that 
M      AB  X  BC 
N  ""  BExBG* 

Place  the  rectaDgles  so  that  the 
angles  ABC  and  GBE  shall  be  opposite, 
i.  e.,  so  that  AG  and  CE  shall  be  straight  Fig.  I73. 

lines  (109) 

Complete  the  rectangle  CBGH,  and  call  it  0. 

Since  M  and  0  have  equal  altitudes, 

-  =  —  .  (1) 

0       BG  ^  ' 

In  like  manner,  since  N  and  0  have  equal  altitudes, 

N_  BE 
0~  BC 


(3) 


Dividing  the  members  of  (1)  by  the  corresponding  members  of  (2), 
we  have 

M      AB  X  BC 

n  =  beTbg-  ^•^^• 


Secon"d  Demonstration. 
Let  ABCD  and  EFGH  be  any  two  rectangles. 


We  are  to  prove  that 


Fig.  174. 

ABCD       AB  ) 


AD 


EFGH       EF  X  EH 

The  bases  and  altitudes  of  the  two  rectangles  are  at  least  to  be  con- 
sidered as  commensurable  by  an  infinitesimal  unit  (342)- 


158 


ELEMENTARY    GEOMETRY. 


Fig.  174. 


Let  i  be  the  common  measure  of  AB,  AD,  EF,  and  EH,  and  suppose  it 
contained  in  AB  m  times,  in  AD  ti  times,  in  EF  p  times,  and  in  EH  q  times. 


AB  AD 

Whence,    m  =  ^  ,    7i  =  -;- 


EF  ^  EH 


Now  conceive  the  rectangles  divided  into  squares  by  drawing  through 
the  points  of  division  of  the  bases  and  altitudes  parallels  to  the  altitudes 
and  bases,  as  in  (336),  whence  the  rectangles  will  be  divided  into  equal 
squares. 

Of  these  equal  squares,  ABCD  contains  m  x  n,  and  EFGH  p^-q. 


Therefore 


ABCD      rnxn 


AB      AD 

— r-    X   — r- 
l  I 


ABxAD 


EFGH  ~  pxg  -  EF      EH        EFxEH 

— :-    X  — r- 
%  % 


Q.  E.  D. 


PROPOSITION    VII. 

344.   Theorem.— The  area  of  a  rectangle  is  equal  to 
the  product  of  its  base  and 
altitude. 


Demonstration. 
Let  ABCD  be  a  rectangle. 

We  are  to  prove  that  its  area 
is  AB  X  AD. 

Let  the  square  u  be  the  pro- 
posed unit  of  measure,  whose  side 
isl. 


Fig.  175. 


\^^V^l 


EQUIVALENCY   AND    AREA.  159 

T,    /«^A^  ABCD       AB  X  AD      .„        ._ 

By  (343),  — ^-  =     ,  =  AB  X  AD. 

Hence,  by  (330),  area  ABCD  =  AB  x  AB.     Q.  E.  D. 

346.  Corollary  1. — The  area  of  a  square  is  equal  to  the 
second  power  of  one  of  its  sides,  as  in  this  case  the  base  and 
altitude  are  equal. 

346.  Corollary  2. — The  area  of  any  parallelogram  is 
equal  to  the  product  of  its  hose  and  altitude  ;  for  any  paral- 
lelogram is  equivalent  to  a  rectangle  of  the  same  base  and 
altitude  (332). 

347.  Corollary  3. — The  area  of  a  triangle  is  equal  to 
one-half  the  product  of  its  base  and  altitude ;  for  a  tnangle 
is  one-half  of  a  parallelogram  of  the  same  base  and  altitude  (333). 

348.  Corollary  4. — Parallelograms  or  triangles  of 
equal  bases  are  to  each  other  as  their  altitudes;  of 
equal  altitudes,  as  their  bases ;  and  in  general  they  are 
to  each  other  as  the  product  of  their  basses  by  their  al^ 
titudes, 

349.  Scholium. — The  arithmetical  signification  of  the  theorem,  The 
area  of  a  rectangle  is  equal  to  the  product  of  its  hose  and  altitude^  is  this : 

Let  the  base  be  6  and  the  altitude  a ;  then  we  have,  by  the  prop- 
osition, 

area  =  ah. 

Now,  in  order  that  ah  may  represent  a  surface,  one  of  the  factors 
must  be  conceived  as  a  surface  and  the  other  as  a  number.  Thus,  we 
may  conceive  h  to  represent  h  superficial  units,  i.  e.,  the  rectangle  having 
the  base  of  the  rectangle  for  its  base  and  being  1  linear  unit  in.altitude. 

The  entire  rectangle  is,  then,  a  times  the  rectangle  which  contains  b 
superficial  units,  or  ah  superficial  units. 

In  the  expression 

area  ABCD  =  AB  x  AD, 

AB  and  AD  may  be  given  a  similar  interpretation. 


160  ELEMENTARY    GEOMETRY, 


PROPOSITION    VIII. 

360.  Theorem. — The  area  of  a  trapezoid  is  equal  to 
the  product  of  its  altitude  into  one-half  the  sum  of  its 
parallel  sides,  or,  what  is  the  same  thing,  the  product 
of  its  altitude  into  a  line  joining  the  middle  points  of 
its  inclined  sides. 

Demonstration^. 

Let  ABCD  be  a  trapezoid,  whose  parallel  sides  are  AB  and  DC,  and 
whose  altitude  is  IK. 

We  are  to  prove,  Ist,  that 

An^n.       ..,       AB  +  CD 
area  ABCD  =  IK  x  — ^ — ' 

and,  2d,  that     area  ABCD  -  \K  x  db,  ^     .^r 

.       '  '  r  ig.  I/O. 

ab  being  a  line  joining  the  middle  points  of  AD  and  BC. 

Draw  either  diagonal,  as  AC.  The  trapezoid  is  thus  divided  into  two 
triangles,  whose  areas  are  together  equal  to  one-half  the  product  of  their 
common  altitude  (the  altitude  of  the  trapezoid)  into  their  bases  DC  and 
AB,  or  this  altitude  into  ^  (AB  +  DC).    Q.  B.  D. 

At  a  and  h  draw  the  perpendiculars  om  and  pn,  meeting  DC,  pro- 
duced, if  necessary. 

Now  the  triangles  aoD  and  A«w  are  equal,  since 

Aa  =  aD, 

angle  o  =  angle  m, 

both  being  right,  and  angle  oaD  =  Aa>n,  being  opposite.     Whence 

f^m  =  oD. 

In  like  manner,  we  may  show  that 

Cp  =  nB. 

Hence,  ab  =  ^(op  +  mn)  (?)  =  ^  (AB  +  DC) ;  and  area  ABCD,  which 
equals  |  (AB  +  DC)  x  IK,  =  a5 x  IK.    Q.  E.  D. 


EQUIVALENCY  AND    AREA, 


161 


PROPOSITION    IX. 

351.  Theorem. — The  area  of  a  regular  polygon  is 
equal  to  one-half  the  product  of  its  apothem  into  its 
perimeter. 

Demonstration. 

Let  ABCDEFG  be  a  regular  polygon,  whose  perimeter  is  AB  +  BC + 
CD  +  DE  +  EF  +  FG  +  GA,  and  whose  apothem  is  Oa, 

We  are  to  prove  that 

area  ABCDEFG  =  iOa(AB  +  BC  +  CD  +  DE  +  EF  +  FG  +  GA). 

Draw  the  inscribed  circle,  the  radii  Oa,  05, 
etc.,  to  the  points  of  tangency,  and  the  radii  of 
the  circumscribed  circle  OA,  OB,  etc.  (273, 
274). 

The  polygon  is  thus  divided  into  as  many 
equal  triangles  as  it  has  sides. 

Now,  the  apothem  (or  radius  of  the  in- 
scribed circle)  is  the  common  altitude  of  these 
triangles,  and  their  bases  make  up  the  perimeter 
of  the  polygon.  *''9-  "77. 

Hence,  the  area  =  ^Oa  (AB  +  BC  +  CD  +  DE  +  EF  +  FG  -I-  GA).     q.  e.  d. 

852.  CoROLLART. — The  area  of  any  polygon  in  which 
a  circle  can  he  inscribed  is  equal  to  one-half  the 
product  of  the  radius  of  the  inscribed  circle  into  the 
perimeter. 

The  student  should  draw  a  figure  and  observe  the  fact.  It  is  espe- 
cially worthy  of  note  in  the  case  of  a  triangle.    See  Fig.  187. 


PROPOSITION    X. 

353.  Lemma. — //  any  polygon  is  circum^scribed 
about  a  circle  and  a  second  polygon  is  formed  by  draw- 
ing tangents  to  the  arcs  intercepted  between  the  con- 
secutive points  of  tangency,  thus  forming  a  polygon  of 
double  the  number  of  sidles,  the  perimeter  of  the  second 
polygon  is  less  than  that  of  the  first. 


162 


ELEMENTARY    GEOMETRY. 


Demonstration. 

Let  ABODE  be  any  circumscribed  polygon,  whose  consecutive  sides 
are  tangent  at  K,  F,  G,  etc.,  and  let  a  second  polygon  be  formed  by 
drawing  tangents  at  f,  g,  etc. 

We  are  to  prove  that  the 
perimeter  ab -^-hc -\- cd^  etc.,  is  less 
than  the  perimeter  EA  +  AB  +  etc. 

Observing  the  portions  of  the 
perimeters  from  K  to  F,  for  the 
first  polygon  we  have 

KA  +  AF  =  Ka  +  (aA  +  Aft)-f5F, 
and  for  the  second 

But    db  <  ak-\-fib  (?). 

Hence, 
Ka  +  ab  +  W  <  KA  +  AF. 

Now,  as  a  similar  reduction 
will  take  place  at  each  vertex,  the  entire  perimeter  of  the  second  polygon 
will  be  less  than  that  of  the  first,     q.  e.  d. 

364.  The  Limit  of  a  varying  quantity  is  a  fixed  quantity 
which  it  approaches  by  such  a  law  as  to  be  capable  of  b.eing 
made  to  differ  from  it  by  less  than  any  assignable  quantity. 

Such  a  varying  quantity  is  often  spoken  of  as  reaching  its 
limit  after  an  infinite  number  of  steps  of  approach. 

365.  Corollary. — As  the  number  of  the  sides  of  a  cir- 
cumscribed regular  polygon  is  increased  the  perimeter 
is  diminished,  andj  approaches  the  circumference  of  the 
circle  as  its  limit,  since  tlie  circle  is  the  limit  of  such  a  poly- 
gon. 


Fig.    178. 


PROPOSITION    XI. 

366.    Tlieorem. — The  area  of  a  circle  is  equal  to  one- 
half  the  product  of  its  radius  into  its  circumference. 

Demonstration. 
Let  Oa  (Fig.  179)  be  the  radius  ofthe  circle. 


EXERCISES. 


163 


"We  are  to  prove  that  the  area  of  the  circle 
is  ^Oa  X  the  circumference. 

Circumscribe  any  regular  polygon. 

Now  the  area  of  this  j^olygon  is  one-half 
the  product  of  its  apothem  and  perimeter. 

Conceive  the  number  of  sides  of  the  poly- 
gon indefinitely  increased,  the  polygon  still 
continuing  to  be  circumscribed  and  regular. 

The  apothem  continues  to  be  the  radius  of 
the  circle,  and  the  perimeter  approaches  the 
circumference. 

When,  therefore,  the  number  of  sides  of  the  polygon  becomes  infinite, 
it  is  to  be  considered  as  coinciding  with  the  circle,  and  its  perimeter  with 
the  circumference  (366). 

Hence  the  area  of  the  circle  is  equal  to  one-half  the  product  of  its 
radius  into  its  circumference,    q.  e.  d. 


Fig.  179. 


367.  A  Sector  is  a  part  of  a  circle  included  between  two 
radii  and  their  intercepted  arc. 

368.  Corollary  1. — The  area  of  a  sector  is  equal  to 
one-half  the  product  of  the  radius  into  the  arc  of  the 
sector. 

369.  Corollary  2. — The  area  of  a  sector  is  to  the  area 
of  the  circle  as  the  arc  of  the  sector  is  to  the  circumfer- 
ence, or  as  the  angle  of  the  sector  is  to  four  right  angles. 


EXERCISES. 


360.  1.  What  is  the  area  in  acres  of  a  triangle  whose  base  is 
75  rods  and  altitude  110  rods  ? 

2.  What  is  the  area  of  a  right-angled  triangle  whose  sides 
about  the  right  angle  are  126  feet  and  72  feet  ? 

3.  If  two  lines  are  drawn  from  the  vertex  of  a  triangle  to  the 
base,  dividing  the  base  into  parts  which  are  to  each  other  as  2,  3, 
and  5,  how  is  the  triangle  divided  ?    How  does  a  line  drawn 


164  ELEMENTARY    GEOMETRY, 

from  an  angle  to  the  middle  of  the  opposite  side  divide  a  tri- 


V 


le 

4.  What  is  the  area  of  the  largest  triangle  which  can  be  in- 
scribed in  a  circle  whose  radius  is  12,  the  diameter  being  one 
side? 

5.  What  is  the  area  of  a  cross  section  of  a  ditch  which  is 
6  feet  wide  at  the  bottom,  9  feet  at  the  top,  and  3  feet  deep  ? 

6.  If  one  of  the  angles  at  the  base  of  an  isosceles  triangle  is 
double  the  angle  at  the  vertex,  how  many  degrees  in  each  ? 


OF      SI  M  I  LARITY. 

361.  The  primary  notion  of  similarity  is  likeness  of  form. 
Two  figures  are  said  to  be  similar  which  have  the  same  shape, 
although  they  may  differ  in  magnitude.  A  more  scientific  defi- 
nition is  as  follows : 

362.  Similar  Figures  are  such  as  have  their  angles  re- 
spectively equal,  and  their  homologous  sides  proportional. 

363.  Homologous  Sides  of  similar  figures  are  those 
which  are  included  between  equal  angles  in  the  respective  figures. 

364.  In  similar  triangles,  the  homologous  sides  are 
those  opposite  the  equal  angles. 

The  student  should  be  careful,  at  the  outset,  to  mark  the  fact  that 
similarity  involves  two  things^  equality  of  angles  and  proportionality 
OP  SIDES.  It  will  appear  that,  in  the  case  of  triangles,  if  one  of  these 
facts  exists,  the  other  exists  also ;  but  this  is  not  so  in  other  polygons. 

365.  Two  figures  are  said  to  be  Mutually  equiangular 
when  each  angle  in  one  has  an  equal  angle  in  the  other,  and 
Mutually  equilateral  when  each  side  in  the  one  has  an  equal 
side  in  the  other. 


SIMILARITY. 


165 


PROPOSITION    I. 

366.    Theorem. — Triangles  which  are  mutually  equi- 
angular are  similar. 

Demonstration. 

Let  ABC  and  DEF  be  two  mutually  equiangular  triangles,  in  which 
A  =  D,  B  =  E,  C  =  F. 

We  are  to  prove  that  the  sides 
opposite  these  equal  angles  are  pro- 
portional, and  thus  that  the  triangles 
possess  both  the  requisites  of  similar- 
ity, viz.,  equality  of  angles  and  pro- 
portionality of  sides. 

Lay  off  on  CA  CD'  =  FD,  and  on 
CB  CE'  =  FE,  and  draw  D'E'. 

Triangle  CD'E'  equals  triangle 
FDE  (?). 

Draw  AE'  and  BD'. 

Pig.    180. 

Since  angle  CE'D'  =  CBA,  D'E'  is  " 

parallel  to  AB  (?),  and  as  the  triangles  D'E'B  and  D'E' A  have  a  common 
base  D'E'  and  the  same  altitudes,  their  vertices  being  in  a  line  parallel 
to  their  base,  they  are  equivalent  (335). 

Now  the  triangles  CD'E'  and  D'E'A,  having  a  common  altitude,  are  to 
each  other  as  their  buses  (348). 


or 


or 


Hence, 

CD'E'       CD' 
D'E'A  -  DA* 

For  like  reason, 

CD'E'       CE' 
D'E'B  -  E  B* 

Whence,  as 

DEB  =  D  E'A, 

CD'       CE' 
DA  ~  E'B 

By  composition, 

CD'                  CE' 

CD'   '   [>'*       CE'  +  E'B 

CD        CE' 

CA     'CB  ' 

.  -        FE 

CA       CB 

166  ELEMENTA  R  Y     GEOMETR  Y. 

In  a  similar  manner,  by  laying  off  ED  and  EF  in  BA  and  BC  respec- 
tively, wenjan  show  that 

FE       ED 
CB  ~  BA  * 

„  FD       FE       ED 

^^^^^'  CA  =  CB^BA-^-^-^- 

367.  Corollary  1. — //  two  triangles  have  two  angles  of 
one  respectively  equal  to  two  angles  of  the  other,  the  tri- 
angles are  similar  (?). 

368.  Corollary  2. — A  transversal  parallel  to  any  side 
of  a  triangle  divides  the  other  sides  proportionally,  and 
the  sides  are  in  the  ratio  of  either  two  corresponding 
segments. 

For  in  the  demonstration  we  have  D'E'  parallel  to  AB,  and 


or 

And  also 
or,  by  alternation, 


CD'       CE' 
DA  ~  E'B' 

CD'       DA 
CE'  ~  E'B  ' 

CD'       CE' 
CA  ~  CB  ' 

CA      CD' 
CB  ~  CE'  ~ 

DA 
E'B 

PROPOSITION    II. 

369.  Theorem. — //  any  iwo  transversals  cut  a  series 
of  parallels,  their  intercepted  segments  are  proportional. 

Demonstration. 

l8t.    Let  OA  and  O'B'  (Fig.  181)  be  any  two  parallel  transversals 
cutting  the  series  of  parallels  ah,  cd,  r/,  yh,  etc. 

We  are  to  prove  that     =-=  =  v?  =  -^ ,  etc. 
oa      af    .  fh 


SIMILARITY. 


167 


Now 
Hence. 


hd         '     df        ' 


=  1.    etc.  (?) 


ac_ce_eg^ 
hd~  df~ fV 

Q.  E.  D. 

2ih  Let  OA  and  OB 
beanynon-paralleltrans- 
versals  cutting  ah,  cd, 
«/>  O^i'9  etc. 

"We  are  to  prove 


Fig.  181. 


ae  _ce_  _eg 


and 


Since  OA  and  OB  are  non-parallel,  they  meet  in  some  point,  as  0. 
Then,  by  (368),  we  have        q^  ^  ^ 

Oc    _    C6 

Od~df' 

Whence,  by  equality  of  ratios,  we  have 

ae  _  M 
bd~df' 


ee 


eg 


Similarly,  we  may  show  that  ^=  tt^  6^c. 

Hence,  also,  by  alternation,  and  by  equality  of  ratios, 

-  =  ^..     ^^  =  '^^    a«d  ^  =  ^,  etc.     Q.E.D. 
ee      df     hd     fh'  eg      fh' 


PROPOSITION    III. 

370.  Theorem. — Conversely  to  Prop.  I,  If  two  triangles 
have  their  corresponding  sides  proportional,  they  are  sim- 
ilar. 

Demonstration. 

AC       OB       BA 


Let  ABC  and  DEF  have 


OF       FE  ~  ED 


168 


ELEMENTARY    GEOMETRY, 


We  are  to  prove  that  ABC  is 
similar  to  DEF. 

As  one  of  the  characteristics  of 
similarity,  viz.,  proportionality  of 
sides,  exists  by  hypothesis,  we  have 
only  to  prove  the  other,  i.  e.,  that 

A  =  D,  C  =  F,  and  B  =  E. 


Make  CD'  = 
parallel  to  AB. 

FD,  and  draw 

D'E' 

Then,  by  (368), 

CA       CB 
CD'  ~  CE' ' 

Fig.  182. 


and  since  by  construction 
and  by  hypothesis 


CD'  =  FD, 


CA 
FD 


CB 
FE' 


CE'  =  FE. 


Again  the  triangles  D'E'C  and  ABC  are  mutually  equiangular,  since  C 
is  common,  angle  CD'E'=  CAB  (?),  and  angle  CE'D'  =  CBA  (?). 


Whence 


CA^ 
CD' 


AB 
D'E' 


But  by  hypothesis  and  construction 


CA 
CD' 


CA 
DF 


AB 
DE 


Hence  D'E'  =  DE,  and  the  triangles  CD'E'  and  DEF  are  equal  (?). 

Therefore  ABC  and  D'E'C  are  similar;  and  as  D'E'C  =  DEF,  ABC  and 
DEF  are  similar,     q.  e.  d. 


371.  Scholium. — As  we  now  know  that  if  two  triangles  are  mutually 
equiangular,  they  are  similar ;  or,  if  they  have  their  corresponding  sides 
proportional,  they  are  similar,  it  will  be  sufficient  hereafter,  in  any  given 
case,  to  prove  either  one  of  these  facts,  in  order  to  establish  the  similarity 
of  two  triangles.  For,  either  fact  being  proved,  the  other  follows  as  a 
consequence. 


SIMILARITY, 


169 


PROPOSITION    IV. 

372.  Theorem. — Two  triangles  which  have  the  sides 
of  the  one  respectively  parallel  or  perpendicular  to  the 
sides  of  the  other,  are  similar. 


Demonstration. 

Let  ABC  and  A'B'C  be  two  triangles  whose  sides  are  respectively 
parallel  or  perpendicular  to  each  other. 

We  are  to  prove  that  tlie  tri- 
angles are  similar. 

Any  angle  in  one  triangle  is 
either  equal  or  supplemental  to  the 
angle  in  the  other  which  is  included 
between  the  sides  which  are  parallel 
or  perpendicular  to  its  own  sides. 
Thus,  A  either  equals  A',  or  A  +  A' 
=  2  right  angles  (294,  295,  296). 

Now,  if  the  corresponding  angles 

are  all  supplemental,  that  is,  if 

A  +  A'  =  2  right  angles, 

B  +  B'  =  2  right  angles, 

and    C  +  C'  =  2  right  angles, 

the  sum  of  the  angles  of  the  two 
triangles  is  6  right  angles,  which  is 
impossible. 

Again,  if  one  angle  in  one  triangle  equals  the  corresponding  angle  in 
the  other,  as  A  =  A',  and  the  other  angles  are  supplemental,  the  sum  is 
4  right  angles  plus  twice  the  equal  angle,  wiiich  is  impossible.  Hence, 
two  of  the  angles  of  one  triangle  must  be  equal  respectively  to  two 
angles  of  the  other.    Therefore  the  triangles  are  similar  (367)-     Q.  e.  d. 


PROPOSITION    V. 

373.  Theorem. — Two  triangles  having  an  angle  in 
one  equal  to  an  angle  in  the  other,  and  the  sides  about  the 
equal  angles  proportional,  are  similar, 

8 


170 


ELEMENT4.HY    GEOMETRY. 


AC 
DF 


CB 
FE 


Demonstration. 

Let  ABC  and  OEF  have  the  angles  C  and  F  equal,  and 

We  are  to  prove  that  ABC  and  DEF  are  similar. 
Make  CD'  equal  to  FD,  and  draw 
D'E'  parallel  to  AB.     Then  is 

angle  CD'E'  =  angle  CAB, 

whence  the  triangles  are  similar  (367), 
and  by  (368), 

AC CB^ 

DC  (=  DF)  ~  CE' ' 

But,  by  hypothesis, 

AC       CB 
DF  ~  FE' 

Whence     CE'  ■=  FE. 

Hence  the  triangle  CD'E'  is  equal  to  the  triangle  FDE.  Now,  CD'E' 
and  ABC  are  mutually  equiangular.  Hence  DFE  and  ABC  are  mutually 
equiangular  and  consequently  similar,     q.  e.  d. 


Fig.  184. 


PROPOSITION    VI. 

374.  Theorem. — In  any  right-angled  triangle,  if  a 
line  is  drawn  from  the  vertex  of  the  right  angle  perpen- 
dicular to  the  hypotenuse : 

1st.  The  perpendicular  divides  the  triangle  into  two 
triangles,  which  are  similar  to  the  given  triangle,  and 
consequently  similar  to  each  other, 

2d.  Either  side  about  the  right  angle  is  a  mean  propor- 
tional between  the  jvhole  hypotenuse  and  the  adjacent 
segment. 

3d.  The  perpendicular  is  a  jnean  proportional  between 
the  segments  of  the  hypotenuse. 

Demonstration. 

Let  ACB  be  a  triangle  right-angled  at  C,  and  CD  a  perpendicular 
upon  the  hypotenuse  AB  ;  then 


STMtLARlTY. 


171 


Ist.  The  triangles  ACD  and  ACB  have  the 
angle  A  common,  and  a  riglit  angle  in  each ; 
hence  they  are  similar  (367).  For  a  like  rea- 
son, CDB  and  ACB  are  similar.  Finally,  as  ACD 
and  CDB  are  both  similar  to  ACB,  they  are 
similar  to  each  other,     q.  e.  d.  ^'9-  '^^• 

2d.  By  reason  of  the  similarity  of  ACD  and  ACB,  we  have 

AD       AC 
AC  ~  AB' 

DB       CB 


and  from  CDB  and  ACB,  we  have 


CB       AB 


Q.  E.  D. 


3d.  By  reason  of  the  similarity  of  ACD  and  CDB,  we  have 


AD 
CD 


CD 

DB"    ^' 


E.  D. 


Queries. — To  which  triangle  does  the  first  CD  belong  ?  To  which 
the  second?  Why  is  CD  made  the  consequent  of  AD?  Why,  in  the 
second  ratio,  are  CD  and  DB  to  be  compared? 

375.  Corollary. — //  a  perpendicular  is  Let  fall  from 
any  point  in  a  circumference  upon  a  diameter,  this  per- 
pendicular is  a  mean  proportional  between  the  segments 
of  the  diameter. 

Let  CD  be  such  perpendicular,  and  draw 
AC  and  CB.  Then,  since  ACB  is  a  right  angle 
(192),  we  have,  by  Case  3d,  the  proportion 

^^§i.    or    CD-AOxDB. 

^"        "°  Fig.  186. 


PROPOSITION    VII. 

376.  Theorem. — The  square  described  on  the  hypote- 
nuse of  a  right-angled  triangle  is  equivalent  to  the  sum; 
of  the  squares  described  on  the  other  two  sides. 

First  Demonstration. 
Let  ACB  (Fig.  187)  be  any  right-angled  triangle. 


m 


ELEMENTARY    GEOMETRY. 


We  are  to  prove  that  AB^  =:  AC^  +  CbI 
For,  let  fall  the  perpendicular  CD,  and  by 
(374,  3d)  we  have 


AD 
AC 


AC 
AB 


v^ ,     and 


DB 
CB 


CB 
AB 


and 


ADxAB  =  AC'; 
DB  X  AB  =  CB  . 


or 


Fig.  187. 


Adding,  we  have   AB  (AD  +  DB)  =  AC^  4-  CB^ 

AB  X  AB  =  AB'  =  AC^+  CB'.     q.  e.  d. 


Second  Demonstration. 
Let  ABC  be  any  right-angled  triangle,  right-angled  at  B. 

Describe  the  squares  AE,  AG,  and  CL  on 
the  hypotenuse  and  the  other  sides  respect- 
ively. From  the  right  angle  let  fall  upon 
DE  the  perpendicular  BK  intersecting  AC  in 
I,  and  draw  the  diagonals  BE,  DB,  HC,  and 
AF. 

Now  the  triangles  BAD  and  HAC  are 
equal,  having  two  sides  and  the  included 
angle  of  one  equal  to  two  sides  and  the  in- 
cluded angle  of  the  other;  viz.,  BA  =  HA, 
being  sides  of  the  same  square,  and  for  a 
like  reason  AD  =  AC ;  and  the  angle  HAC 
=  BAD,  since  each  is  made  up  of  a  right 
angle  and  the  angle  BAC. 

Since  ABG  and  ABC  are  right  angles,  BG  is  the  prolongation  of  BC, 
and  the  triangle  HAC  has  the  same  base,  HA,  and  the  same  altitude,  AB, 
as  the  square  AG.     Hence  the  triangle  HAC  is  half  the  square  AG. 

Moreover,  the  triangle  BAD  has  the  same  base,  AD,  as  the  rectangle 
AK,  and  the  same  altitude  as  Al.     Hence, 

triangle  BAD  =  ^ADKI. 

Therefore,  as  the  rectangle  ADKI  and  the  square  AG  are  twice  the 
equal  triangles  BAD  and  HAC  respectively,  they  are  equivalent. 

In  like  manner,  the  square  CL  may  be  shown  to  be  equivalent  to  the 
rectangle  CK. 

Whence  we  have  ADKI  =  ABGH, 

and  IKEC  ^  BCFL; 

and  adding,  ADEC  =  ABGH  f  BCFL.     Q.  e.  d. 


Fig.  188. 


SIMILARITY.  173 

377.  Corollary  1. — The  hypotenuse  of  a  right-angled 
triangle  equals  the  square  root  of  the  sum  of  the  squares 
of  the  other  two  sides. 

Also,  either  side  about  the  right  angle  equals  the  square 
root  of  the  square  of  the  hypotenuse  minus  tJie  square  of 
the  other  side. 

378.  Corollary  2. — The  diagonal  of  a  square  is  a/2 
times  the  side. 

For,  let  S  be  the  side.  Drawing  the  diagonal,  we  have  a  right-angled 
triangle  of  which  the  diagonal  is  the  hypotenuse,  and  the  sides  about  the 
right  angle  are  each  S.     Hence,  by  the  proposition, 

(diag.)«  =  S'  +  8'  =  25«, 
or  diag.  =  8^'i. 

379.  Scholium. — Proposition  VI  with  its  corollary,  and  Prop.  VII, 
which  is  a  direct  result  of  Prop.  VI,  are  perhaps  the  most  Iruitful  in 
direct  practical  results  of  any  in  Geometry.  Prop.  VII  is  called  the 
Pythagorean  Proposition,  its  original  demonstration  being  attributed  to 
Pythagoras. 


PROPOSITION    VIII. 

380.  Theorem. — Regular  polygons  of  the  sam,e  num- 
ber of  sides  are  similar  figures. 

Demonstration. 

Let  P  and  P'  be  two  regular  polygons  of  the  same  number  of  sides, 
a,  b,  c,  d,  etc.,  being  the  sides  of  the  former,  and  a',  h' ,  c' ,  d' ,  etc., 
the  sides  of  the  latter. 

Now,  by  the  definition  of  regular  polygons,  the  sides  a,  ft,  c,  d,  etc., 
arc  equal  each  to  each,  and  also  a\  b\  c',  d\  etc.     Hence,  we  have 
a^  _  b  _  c  _  d 
a'"  V  ~  d  ~  W       ' 
Again,  the  angles  are  equal,  since  n  being  the  number  of  angles  of 
each  polygon,  each  angle  is  equal  to 


174  ELEMENTARY    GEOMETRY. 

n  X  2  right  angles  —  4  right  angles  (norj\ 
n 
Hence  the  polygons  are  mutually  equiangular,  and  have  their  corres- 
ponding sides  proportional ;  that  is,  they  are  similar,     q.  e.  d. 


PROPOSITION    IX. 

381.  Theorem. — The  corresponding  diagonals  of  reg- 
ular polygons  of  the  same  number  of  sides  are  in  the  same 
ratio  as  the  sides  of  the  polygons. 

[Let  the  student  give  the  demonstration.] 


PROPOSITION    X. 

382.  Theorem. — TJie  radii  of  the  circumscribed,  and 
also  of  the  inscribed  circles,  of  regular  polygons  of  the  same 
numjber  of  sides,  are  in  the  saine  ratio  as  the  sides  of  the 
polygons, 

Demonstkatioij. 


Fig    189. 

Let  ABCDEF  and  abcdefbe  two  regular  polygons  of  the  same  num- 
ber of  sides,  and  It  and  r  be  the  radii  of  their  circumscribed  circles, 
and  II'  and  r'  of  their  inscribed. 

We  are  to  prove  that  —  ( =  — ,  etc.  J  z=  —  =  —  - 


SIMILARITY.  176 

Let  0  and  0'  be  the  centres  of  the  polygons,  and  draw  OA,  OF,  O'a, 
and  0/,  and  also  the  apothenis  01  and  O'i. 

OA  =  R,        and         O'a  =  r  (?); 
also  01  =  /?,        and        O'i  =  r'   (?). 

Now  the  triangles  AFO   and  nfO'  are  equiangular  (?),   and  hence 

similar. 

rru      r  AF/     FE     ^   \      OA      i? 

Therefore,  —-(  =  —-,  etc.  )  =  -r-  =  - .    q.  e.  d. 

a/  V     /«  J      O'a       r     ^ 

Again,  the  triangles  AlO  and  aiO'  are  mutually  equiangular  (?),  and 
hence  similar. 


Therefore,  _ .  —  ,rvr»> 


Ai_qi 

ai  ~  or 

whence,  doubling  the  terms  of  the  first  ratio,  we  have 
^F/      F£     ^   \      0\      B' 


383.  Homologrous  Altitudes  in  similar  triangles  are 
perpendiculars  let  fall  from  the  vertices  of  equal  angles  upon  the 
sides  opposite. 

384.  Homologfous  Diagonals  in  similar  polygons  are 
diagonals  joining  the  vertices  of  corresponding  equal  angles. 


PROPOSITION    XI. 

385.    Theorem.^-Hojnologous  nZtitivdes  in  simUar  trU 
angles  have  the  same  ratio  as  the  homologous  sides. 
[Let  the  student  give  the  demonstration.] 


PROPOSITION    XII. 

386.  Theorem. — The  bisectors  of  equal  angles  of  simi^ 
lar  triangles  are  to  each  other  as  the  homologous  sides  of 
the  triangles,  hence  as  the  homologous  perpendiculars, 

[Let  the  student  ^ve  the  demonstration,] 


176 


ELEMENTARY    GEOMETRY. 


PROPOSITION     XIII. 

387.    Theorem. — Homologous    diagonals    in   similar 
polygons  have  the  same  ratio  as  the  homologous  sides. 

Demonstration. 

Let  ABCDEFG  and  ahcdefg  be  two  similar  polygons,  having  angle 
A  =  angle  a,  B  =  6,  C  =  c,  etc. 


Fig.   190. 


We  are  to  prove  that 


AC  AD      ^ 

— ,  or  — 3  ,  etc. 
oc  '         ad 


AB 


AB 


the  ratio  — r^  being  the  ratio  of  any  two  homologous  sides  of  the  polygons. 
The  triangles  ABC  and  dbc  are  similar  (?),  and  hence 

AC^AB 

ac  ~  ab 

Also,  since  triangle  ABC  is  similar  to  a6c, 

angle  BCA  =  angle  Jm, 

and  subtracting  these  respectively  from  the  equal  angles  (?)  BCD  and  Icd^ 
we  have 

angle  ACD  =  angle  acd. 

Hence  the  two  triangles  ACD  and  acd  have  an  angle  in  each  equal 
and  the  including  sides  proportional  (?),  and  are  consequently  similar. 

AD  ^  AC  ^  AB 

ad        ac  ~  ab 


Therefore 


SIMILARITY. 


177 


In  like  manner,  any  homologous  diagonals  may  be  shown  to  have  the 
AB 


ratio  '-^^ ,  which  is  the  ratio  of  any  two  homologous  sides. 
ab  ' 


Q.  E.  D. 


388.  Corollary  1. — Any  two  similar  polygons  are  di- 
vided by  their  homologous  diagonals  into  an  equal  number 
of  similar  triangles  similarly  placed, 

389.  Corollary  2.— Conversely,  Tivo  polygons  which  can 
be  divided  by  diagonals  into  the  same  number  of  mutually 
similar  triangles,  similarly  placed,  are  similar. 


PROPOSITION    XIV. 
390.    Theorem. — Circles  are  similar  figures. 


Fig.  191. 


Demonstration. 
Let  0«  and  OA  be  the  radii  of  any  two  circles. 

Place  the  circles  so  that  they  shall  be  con- 
centric, as  in  the  figure.  Inscribe  the  regular 
hexagons,  as  ahcdef,  ABCDEF. 

Conceive  the  arcs  AB,  BC,  etc.,  of  the  outer 
circumference  bisected,  and  the  regular  do- 
decagon inscribed,  and  also  the  corresponding 
regular  dodecagon  in  the  inner  circumference. 

These  are  similar  figures  by  (380). 

Now,  as  the  process  of  bisecting  the  arcs 
of  the  exterior  circumference  can  be  conceived 
as  indefinitely  repeated,  and  the  corresponding  regular  polygons  as  in- 
scribed in  each  circle,  the  circles  may  be  considered  as  regular  polygons 
of  the  same  number  of  sides,  and  hence  similar,     q.  e.  d. 

391.  Corollary. — Sectors  which  correspond  to  equal 
angles  at  the  centre  are  similar  figures. 

Since  a  radius  is  perpendicular  to  the  circumference  of  its  circle,  such 
sectors  are  mutually  equiangular ;  and  by  the  proposition  it  is  evident 
that  the  arcs  are  to  each  other  as  the  radii. 

i  e     ^^^^  =  ^ 
'    '  arc  FE       OF' 

Scholium. — The  circle  is  said  to  be  the  limit  of  the  inscribed  polygon, 
and  the  circumference  the  limit  of  the  perimeter.     By  this  is  meant  that 


178  ELEMENTARY    GEOMETRY, 

as  the  number  of  the  sides  of  the  inscribed  polygon  is  increased  it  ap- 
proaches nearer  and  nearer  to  equality  with  the  circle.  The  apothem 
approaches  equality  with  the  radius,  and  hence  has  the  radius  for  its 
limit. 


PROPOSITION    XV. 

392.  Problem.— To  divide  a  given  line  into  parts 
which  shall  be  proportional  to  several  given  lines. 

Solution.* 

Let  it  be  required  to  divide  OP  into  parts  pro- 
portional to  the  lines  A,  B,  C,  and  D. 

Draw  ON  making  any  convenient  angle  with 
OP,  and  on  it  lay  off  A,  B,  0,  and  D,  in  succession, 
terminating  at  M. 

Join  M  with  the  extremity  P,  and  draw  par- 
allels to  MP  through  the  other  points  of  division. 

Then  by  reason  of  the  parallels  we  shall  have 
A:B:C:D    ::    a  :  h  :  c  :  d  (369). 

Fig.  192. 

393.  The  notation  A  :  B  :  C  :  D  : :  a  :  5  :  c  :  c^  is  of  such  frequent 
occurrence  in  mathematical  writing  that  we  feel  constrained  to  retain  it. 
It  means  that  the  successive  ratios 

ABC 
BC     D' 

are  equal  to  the  successive  ratios 

a      b       c 
b'     c/     d' 

We  may  read  the  expression  thus :  "  The  successive  ratios  A  to  B, 
B  to  0,  0  to  D  =  the  successive  ratios  a  to  b.b  to  c,  c  to  <Z."  It  does  rwt 
mean  that  the  ratio  A  to  B  =  B  to  0,  etc. 

*  Hereafter  we  shall  change  somewhat  the  style  of  our  demonstrations, 
from  the  elementary  form  hitherto  used  to  the  more  common  and  free  form 
used  by  writers  generally.  In  the  "  Solution  "  of  a  problem  we  shall  here- 
after usually  include  the  "  Demonstration  of  the  Solution." 


siMiLAEirr, 


179 


PROPOSITION    XVI. 

394.    Problem,— To  find  a  fouHh  proportional  to  three 
given  lines. 

Solution. 

Let  it  be  required  to  find  D,  a  fourth  proportional  to  the  lines  A,  B, 
and  C,  so  that  we  shall  have  g  =  q  ' 


From  8(1111  e  point  0,  draw  two 
indefinite  lines  OX,  OY.  Lay  off  on 
OX,  Oa  =  A,  and  Oc  =  B.  Also,  on 
OY  lay  off  06  =  C,  and  draw  (th. 
Through  c  draw  cd  parallel  to  oh. 
Then  is  Od  the  fourth  proportional, 
D,  which  was  sought. 

For,  since  ah  and  cd  are  parallel, 
we  have,  by  (368), 

Oa  (or  A)  _  Oh   or  0)  ^ 
Oc  (or  B)  ~  Od  (or  D)  ' 

Hence  D  is  the  fourth  proportional  sought. 


Fig.  193. 


395.  Scholium. — In  speaking  of  the  fourth  proportional  to  three 
given  lines,  it  is  necessary  that  the  order  in  which  the  three  are  to  occur 
be  specified.  This  order  is  usually  understood  to  be  that  in  which  the 
lines  are  named.  Thus,  a  fourth  proportional  to  A,  B,  and  0,  is  D,  as 
found  above.  But  a  fourth  proportional  to  B,  A,  and  C  is  quite  a  differ- 
ent line  from  D. 


PROPOSITION    XVII. 

396.    Problem. — To  find  a  third  proportional  to  two 
given  lines. 

Solution. 


Let  A  and  B  be  the  two  given  lines. 


180 


ELEMENTARY    GEOMETRY, 


We  are  to  find  a  third  proportional, 
«,  such  that 

A       B 
B~  x' 

The  usual  solution  is  the  same  as  the 
last,  C  being  equal  to  B.  [Let  the  stu- 
dent execute  it.]  Fig.  I94. 

Another  Solution. 
Let  A  and  B  be  the  two  lines. 

Draw  an  indefinite  line  AM,  and  take 
AD  =  A 

At   D  erect  a    perpendicular   BD   and 
make  it  equal  to  B. 

Join  A  and  B,  and  bisect  it  by  the  per- 
pendicular ON. 

NO  will  intefsect  AM ;  since,  as  A  is  less  Fig-  '95. 

than  a  right  angle  (?),  the  sum  of  the  two  angles  ONA  and  OAN  is  less 
than  two  right  angles  (129). 

From  0  as  a  centre,  with  OA  as  a  radius,  describe  a  semi-circumfer- 
ence.    It  will  pass  through  B  (?). 


Now 


AD  (or  A)       BD  (or  B) 


(?). 


BD  (or  B)       DC  (or  x) 
Hence,  CD  =  a;,  the  required  third  proportional. 


PROPOSITION    XVIII. 

397.    Problem. — To  find  a  mean  proportional  between 
two  given  lines. 

Solution. 


Let  it  be  required  to  find  a  mean 
proportional,  .r,  between  M  and  N,  so 
that 

M  _  « 

a;  ~  N 


x=  ^/Mx  "N. 


Fifl.  196. 


EXERCISES.  181 

Draw  an  indefinite  line,  and  on  it  lay  off  AD  =  M,  and  DB  =  N.  On 
AB  as  a  diameter  draw  a  semi-circumference,  and  erect  DC  perpendicular 
to  AB.     Then  CD  =  x,  the  mean  proportional  required. 

[Let  the  student  give  the  proof.] 


PROPOSITION    XIX. 

I.    Problem. — To  construct  a  square  equivalent  to  a 
given  tri angle. 

Find  a  mean  proportional  between  the  altitude  and  half  the  base.     On 
this  construct  a  square. 

[Let  the  student  execute  the  problem  and  demonstrate  it.] 


EXERCISES. 

399.  1.  Draw  any  line,  and  divide  it  into  3,  5,  8,  or  10  equal 
parts. 

2.  Draw  any  line  and  divide  it  into  parts  which  shall  be  to 
each  other  as  2,  3,  and  5. 

3.  Construct  the  square  root  of  7,  11,  2. 
Fig.  197  will  suggest  the  construction 

of  y/1. 

4.  The  diameter  of  a  circle  is  20. 
What  is  the  perpendicular  distance  to 
the  circumference  from  a  point  in 
the  diameter  15  from  one  extremity  ?  ^'fl-  '^^• 

What  are  the  distances  from  the  point  where  this  perpendicular 
meets  the  circumference  to  the  extremities  of  the  diameter  ? 

5.  The  sides  of  one  triangle  are  7,  9,  and  11.  The  side  of  a 
second  similar  triangle,  homologous  with  side  9,  is  A^.  What 
are  the  other  sides  of  the  latter  ? 


182 


ELEMENTARY    GEOMETRY. 


6.  DE  being  parallel  to  BC,  prove  that  the  tri- 
angles DOE  and  BOC  are  similar,  and  hence  that 

OD       OE 
OC  ~  OB 

Are  the  following  proportions  true  ? 
OD       OE  OD       00 

00  ~  OB  DE  ~  BC' 

OD  _  OC  OB       OE 

OE     ■  BC         BC  "  DE  Fig.  198. 

7.  Draw  any  triangle  or  polygon,  and  then  construct  a  similar 
one  whose  homologous  sides  shall  be  f  as  long. 

8.  Show  that  if  ABCDEF  is  a  regular 
polygon,  khcdef  is  also  regular,  he,  cd,  etc., 
being  parallel  to  BC,  CD,  etc.  Show  that 
any  two  similar  polygons  may  be  placed 
in  similar  relative  positions,  and  hence 
show  that  the  corresponding  diagonals  are 
in  the  same  ratio  as  the  homologous  sides. 

Fig.  199. 


PROPOSITIONS  FOR  ORIGINAL  INVES- 
TIGATION. 


400.  1.  If  two  straight  lines  join 
the  alternate  ends  of  two  parallels, 
the  line  joining  their  centres'  is  half 
the  difference  of  the  parallels. 

We  are  to  prove  that 

EF  =  i(CD- AB). 
^CH  =  EF  =  HCD-AB). 


Fig.  200. 


2.   To  construct  a  square  equivalent  to  a  given  polygon. 

First  reduce  the  polygon  to  a  triangle  (339).     Then  construct  an 
equivalent  square  (398)- 


EXERCISES.  183 

3.  The  area  of  a  regular  inscribed  dodecagon  is  three 
times  the  square  on,  the  radius. 

4.  //  the  sides  of  a  quadrilateral  he  divided  into  m 
equal  parts,  and  the  ii^^  points  of  division,  reckoning  from 
two  opposite  vertices,  he  joined  so  as  to  form  a  quadri- 
lateral, the  quadrilateral  will  he  a  parallelogram,. 


Fig  201. 

5.  The  line  drawn  from  the  vertex  of  the  right  angle 
of  a  right-angled  triangle  to  the  middle  of  the  hypotenuse 
is  half  the  hypotenuse. 

Prove  from  either  figure. 

6.  In  any  triangle  the  rectangle  of  two  sides  is  equiva- 
lent to  the  rectangle  of  the  perpendicular 
let  fall  from  their  included  angle  upon  the 
third  side,  into  the  diameter  of  the  circum- 
scribed circle. 

This  proposition  is  an  immediate  consequence  ot 
the  similarity  of  two  triangles  in  the  figure.  pjg^  202! 


184  ELEMENTAR  Y     GEOMETB  Y, 


APPLICATIONS  OF  THE  DOCTRINE  OF  SIMILARITY  TO 
THE  DEVELOPMENT  OF  GEOMETRICAL  PROPERTIES 
OF  FIGURES. 

401.  The  doctrine  of  similarity,  as  presented  in  the  preceding 
section,  is  the  chief  reliance  for  the  development  of  the  geomet- 
rical properties  of  figures.  This  section  will  be  devoted  to  the 
investigation  of  a  few  of  the  more  elementary  properties  of  plane 
figures,  which  we  are  able  to  discover  by  means  of  this  doctrine. 


OF    THE    RELATIONS 

OFTHE  SEGMENTS  OF  TWO  LINES  INTERSECT- 
ING EACH  OTHER,  AND  INTERSECTED  BY  A 
CIRCUMFERENCE. 


PROPOSITION    I. 

402.    Theorem. — //  two  chords  intersect  each  other  in 

a  circle,  •  their  segments  are  reciprocalhj  proportioned ; 
whence  the  product  of  the  segments  of  one  chord  equals 
the  product  of  the  segments  of  the  other. 

Demonstration. 
Let  the  chords  AC  and  BD  (Fig.  203)  intersect  at  0. 

We  are  to  prove  that  OA  ~  OD  * 

whence  OB  x  OD  =  OA  x  OC. 


APPLICATIONS   OF  DOCTRINE   OF  SIMILARITY, 


185 


Draw  AD  and  BC. 

The  two   triangles   AOD  and   BOC  are  simi- 
lar (?). 

OB     qc 

OA  ~  OD' 

OB  X  OD  =  OA  X  00.     Q.  E.  D. 


Hance, 
whence 


Queries. — Why   is   OB  compared   with  OA? 
Why  00  with  OD  ?     Would  AO  :  CO  : :  BO  :  DO  Fig.  203. 

be  true  ?    Would  AO  :  DO  : :  BO  :  CO  ?    What  is  the  force  of  the  word 
"  reciprocally,"  as  used  in  the  proposition  ? 


PROPOSITION    II. 

403.  Theorem. — //  from>  a  point  ivithout  a  circle,  two 
secants  are  draivn  terminating  in  the  concave  arc,  the  whole 
secants  are  reciprocally  proportional  to  their  external  seg- 
ments ;  whence  the  product  of  one  secant  into  its  external 
segm^ent  equals  the  product  of  the  other  into  its  external 
segment. 

Demonstration. 

Let  OA  and  OB  be  two  secants  intersecting  the  circumference  in  D 
and  C  respectively. 

We  are  to  prove 

OB  _  OD 
OA  ~  OC ' 

whence,         OB  x  OC  =  OA  x  OD. 

Draw  AC  and  BD. 

The  two  triangles  AOC  and  BOD  are  simi- 
lar (?). 

OB       OD 

OA  ~  OC ' 


Hence, 


whence,         OB  x  OC  =  OA  x  OD.     q.  e.  d. 


Fig.  204. 


Queries.  —Same  as  under  preceding  demonstration. 


186 


ELEMENTARY    GEOMETRY, 


PROPOSITION    III. 

404.  Theorem. — //  from  a  point  without  a  circle  a 
tangent  is  drawn,  and  a  secant  terminating  in  the  con- 
cave arc,  the  tangent  is  a  mean  proportional  between  the 
whole  secant  and  its  external  segment ;  whence  the  square 
of  the  tangent  equals  the  product  of  the  secant  into  its 
external  segment. 

Demonsteation. 


Let  OA  be  a  tangent  and  OB  a  secant  intersecting  the  circumference 


in  C. 

We  are  to  prove  that 

OB       0  A 

OA     be  * 

whence,  OB  x  OC  =  OA  . 

Draw  AC  and  AB. 

The  two  triangles  AOB  and  AOC  are  simi' 
lar,  since  angle  0  is  common,  and  angle  OAC  — 
angle  B  (?). 

„  OB       OA 

Hence, 


Fig.  205. 


whence, 


OA       00* 

OB  X  00  =  OAI     q.  e.  d. 


OF  THE    BISECTOR    OF   AN    ANGLE    OF 
A    TRIANGLE. 


PROPOSITION    IV. 

405.  Theorem. — A  line  which  bisects  any  angle  of  a 
triangle  divides  the  opposite  side  into  segments  propor- 
tional to  the  adjacent  sides. 

Demonstration. 
In  the  triangle  ABC  (Fig.  206)  let  CD  bisect  the  angle  ACB. 


APPLICATIONS    OF    DOCTRINE    OF    SIMILARITY, 
Then  is 


187 


AD       AC* 
DB  ~  CB 


Draw  BE  parallel  to  CD,  and  produce  it 
till  it  meets  AC  produced  in  E. 

By  reason  of  the  parallels  CD  and  EB, 
angle  ACD  =  AEB, 
and  DCB  =  CBE. 

But,  by  hypothesis,  ACD  =  DCB. 

Therefore,  AEB  (or  CEB)  =  CBE, 

and  CE  =  CB  (?). 

AD  _         AC 

DB  ~  CE(=  CB) 


Hence,  finally, 


Fig.  206. 


(368).      Q.  E.  D. 


PROPOSITION    V. 

406.  Theorem. — //  a  line  is  drawn  from  any  vertex 
of  a  triangle  bisecting  the  exterior  angle  and  intersecting 
the  opposite  side  produced,  the  distances  from  the  other 
vertices  to  this  intersection  are  proportional  to  the  adjacent 
sides. 

Demonstration. 


Let  CD  bisect  the  exterior  angle  BCF  of  the  triangle  ACB. 

AD  _  AC 
BD  ~  CB' 


Then  is 


Fig.  207. 


For,  draw   BE   parallel  to  AC. 
By  reason  of  these  parallels, 

angle  FCE  =  CEB, 
and  BCE  =  FCE,  by  hypothesis. 

Hence,  CEB  =  BCE, 

and  CB  =  BE. 

Also,  by  reason  of  the  similar  triangles  ACD  and  BED, 

AD  AC 

BD  ~  BE(orCB) 


Q.  E.  D. 


*  See  note  at  the  bottom  of  p.  178, 


188 


ELEMENTARY    GEOMETRY. 


PROPOSITION    VI. 

407.  Theorem. — //  a  line  is  drawn  bisecting  any 
angle  of  a  triangle  and  intersecting  the  opposite  side,  the 
product  of  the  sides  about  the  bisected  angle  equals  the 
product  of  the  segments  of  the  third  side,  plus  the  square 
of  the  bisector. 

Demokstration. 

In  the  triangle  ACB,  let  CD  bisect  the  angle 
ACB. 

Then  AC  x  CB  =  AD  x  DB  +  CD'. 

For,  circumscribe  the  circle  about  the  trian- 
gle, produce  the  bisector  till  it  meets  the  circum- 
ference at  E,  and  draw  EB.  The  triangles  ADC 
and  CBE  are  similar,  since  angle  ACD  =  ECB,  by 
hypothesis,  and  A  =  E,  because  each  is  measured 
by  ^  arc  CB. 

AC       CD 

CE  ~  CB  ' 


Fig.  208. 


Therefore, 


whence,  ACxCB  =  CE  xCD  =  (DE  +  CD)  CD 

=  DE  X  CD  +  CdI 
For  DE  X  CD,  substituting  its  equivalent  AD  x  DB  (402),  we  have 
AC  X  CB  =  AD  X  DB  +  CD^.     Q.  e.  d. 


AREAS    OF    SIMILAR    FIGURES. 


PROPOSITION    VII. 

408.    Theorem. — The  areas  of  similar  triangles  are  to 

each  other  as  the  squares  described  on  their  homologous 

sides. 

Demonstration. 

Let  ABC  and  EFG  be  two  similar  triangles,  the  homologous  sides 
being  AB  and  EF,  BC  and  FG,  and  AC  and  EG. 


APPLICATIONS    OF   DOCTRINE    OF   SIMILARITY. 


189 


Then  is 

area  ABC      AC^ 


AT      BC=^ 


area  EFG       e^^      e  F^      FG^' 

From  the  greatest  *  angle  in  each  tri- 
angle let  fall  a  perpendicular  upon  the 
opposite  side.  Let  these  perpendiculars 
be  BD  and  FH. 


Now 


and 


BD  _  AC  (.. 
FH-EG^'^' 

jAC  _  AC  ,as 
iEG"EG^^* 


Fig.  209. 


Multiplying  the  corresponding  ratios 

together,  we  have  

iAC^BD       AC^ 
iEGxFH  "  eg' 


and 


But 

Hence, 

And,  finally,  as 


^AC  X  BD  =  area  ABC, 

iEG  X  FH  ^  area  EFG  (?). 

area  ABC  _  AC^ 
area  EFG  ~  eq^ 


AC 


AB'       BC" 


we  have 


EG        EF' 


area  ABC       AC 


area  EFG 


EG" 


FG' 

AT 
EF^ 


(?), 


Bc; 

FG^ 


409. 


PROPOSITION    VIII. 
Theorem. — The  areas  of  similar  polygons  are  to 


each  other  as  the  squares  of  any  two  Jwmologous  sides  of 
the  polygons. 


*  The  only  object  in  taking  the  largest  angles  is  to  make  the  perpendic- 
ular fall  within  the  triangle.  The  demonstration  is  essentially  the  same 
when  the  perpendiculars  fall  upon  the  opposite  sides  produced. 


190 


ELEMENTARY    GEOMETRY. 


Demonstration. 

Let  ABCDEF  and  abcdef  be  two  similar  polygons,  the  homologous 
sides  being  AB  and  a&,  BC  and  6c,  CD  and  cd,  DE  and  de,  EF  and  ef, 
FA  and  /«. 

Let    area  ABCDEF  =  P, 
and  area  abcdef  =  p. 

Then  is 


BC; 

67 


Fig.  210. 


or  as  the  squares  of  any  two 
homologous  sides. 

Draw  the  homologous  di- 
agimals  AC,  AD,  AE,  and  ac,  ad^  and  ae,  dividing  the  polygons  into  the 
similar  triangles  M  and  m^  N  and  w,  0  and  o.  and  S  and  s  (388)* 

r2 


Now 


-  -  ^^  (?) 


ED" 


-de' 


S^CB' 


But 


whence, 


cb; 

a;' 


Dc; 

dc 
=  - 


=  %-%i^y, 


ed 
0 

0 


fe 


Taking  this  by  composition,  we  have 


M+N+O+S 

m+n+o+8 


CB 

And  as  the  ratio  ^  is  the  same  as  that  of  the  squares  of  any  two 

cb' 
homologous  sides,  P  and  p  are  to  each  other  as  the  squares  of  any  two 
homologous  sides. 


APPLICATIONS    OF   DOCTRINE    OF    SIMILARITY.  191 

Finally,  as  this  argument  can  be  extended  to  the  case  of  any  two 
similar  polygons,  the  areas  of  any  two  similar  polygons  are  to  each  other 
as  the  squares  of  any  two  homologous  sides  of  the  polygons,     q.  e.  d. 

410.  Corollary  1. — Similar  polijgons^  are  to  each  other 
as  the  squares  of  their  corresponding  diagonals. 

In  the  demonstration  we  have      -  =  —  =  — -. 

P      m       cb^ 

By  (388,  408)  we  have     ^=^1-^-^ 
^       ae^      adr       a<r 

P      AE2 

Hence  -  =  ^=i ,  etc. 

P      ai 

411.  Corollary  ^.—Regular  polygons*  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  their 
homologous  sides.    [They  are  similar  figures  (?)]. 

412.  Corollary  3. — Regular  polygons  of  the  same  num- 
ber of  sides  are  to  each  other  as  the  squares  of  their 
apothems. 

For  their  apothems  are  to  each  other  as  their  sides.  Hence  the 
squares  of  their  apothems  are  to  eacli  other  as  the  squares  of  their  sides. 

413.  Corollary  4. — Circles  are  to  each  other  as  the 
squares  of  their  radii  (390),  and  as  the  squares  of  their 
diameters. 


OF    PERIMETERS     AND    THE    RECTIFI- 
CATION  OF  THE  CIRCUMFERENCE. 

414.  The  Rectification  of  a  curve  is  the  process  of  find- 
ing its  length. 

The  term  rectification  signifies  making  straight,  and  is  applied  as 
above,  under  the  conception  that  the  process  consists  in  finding  a  straight 
line  equal  in  length  to  the  curve. 


*  This  is  a  common  elliptical  form  for  "  The  a/reas  of,  etc" 


192  ELEMENTARY    GEOMETRY. 


PROPOSITION    IX. 

415.  Theorem. — Tlie  perimeters  of  similar  polygons 
are  to  each  otlier  as  tJieir  homologous  sides,  and  as  their 
corresponding  diagonals. 

Demo:n^stration. 

Let  (If  b,  c,  dy  etc.,  and  A,  B,  C,  D,  etc.,  be  the  homologous  sides  of 
two  similar  polygons  whose  perimeters  are^>  and  P. 

and  r  and  R  being  corresponding  diagonals, 


p  _  r 
P  ~  R 


Since  the  polygons  are  similar, 
By  composition, 


a       h        c       d       ^ 


a  +  h-\-c  +  d4- etc.  (or p)    _  a 
A  +  B  +  C  +  D  +  etc.  (orP)  ~  A    . 

or  as  any  other  homologous  sides.     Also,  as  the  homologons  sides  are  to 
each  other  as  the  corresponding  diagonals  (387), 

p      r 

P       R      ^ 

416.  Corollary  1. — The  perimeters  of  regular  polygons 
of  the  same  number  of  sides  are  to  each  other  as  the  apo- 
thems  of  the  polygons  (382). 

417.  Corollary  2. — The  circmnferences  of  circles  are 
to  eojch  other  as  their  radii,  and  as  their  diam^eters  (390). 


PROPOSITION    X. 

418.  Problem. — To  find  the  relation  between  the 
chord  of  an  arc  and  the  chord  of  half  the  arc  in  a  circle 
whose  radius  is  v. 


APPLICATIONS    OF   DOCTRINE    OF    SIMILARITY. 


193 


Solution. 

Let  0  be  the  centre  of  the  circle,  AB  any  chord,  and  CB  the  chord 
of  half  the  arc  AB. 

Let  AB  =  C,  and  CB  =  c. 

We  are  to  find  the  relation  between  C  and  c. 
Draw  the  radii  CO  and  BO,  and  call  each  r. 
CO  is  perpendicular  to  AB  (?). 
In  the  right-angled  triangle  BDO, 


DO  =  VBO'-iC-^  (?), 
or  DO  =  V*-'  -  iC". 


Fig.  211. 


Hence, 


CD  =  r  -  ^/r^' 


(P, 


Again,  in  the  right-angled  triangle  CDB, 

CB  =  Vcb'  +  bd' 


=  y  2r»  -  2rv^r»  — ;^(7» 
=  |/2r 


iC« 


.y'4r»  -  G\ 


Therefore,  e  =  y  2r'  —  r>v/4r'»  —  C  is  the  relation  desired. 
419.  ScHOLroM. — The  formula 


=  V  2r'-rV'4r*- 


(7« 


is  the  value  of  the  chord  of  half  the  arc  in  terms  of  the  chord  of  the 
whole  arc  and  the  radius.     From  this  we  readily  obtain 


C  =  -  V4r'  -  c\ 

which  is  the  value  of  the  chord  in  terras  of  the  chord  of  half  the  arc  and 
the  radius. 


T8I4  ELEMENTARY    GEOMETRY. 


PROPOSITION    XI. 

420.  Theorem. — The  circumference  of  a  circle  whose 
radius  is  1,  is  2-rT,  the  numerical  value  of  n  being  approj^i- 
matelyS.Uie. 

Demonstration. 

We  will  approximate  the  circumference  of  a 
circle  whose  radius  is  1,  by  obtaining,  1st,  the 
perimeter  of  the  regular  inscribed  hexagon ;  2d, 
the  perimeter  of  the  regular  inscribed  dodeca- 
gon ;  3d,  the  perimeter  of  the  regular  inscribed 
polygon  of  24  sides ;  then  of  48,  etc. 

By  varying  the  polygon  in  this  manner,  it  is 
evident  that  the  perimeter  approaches  the  cir- 
cumference as  its  limit  (282,  354),  since  at  each  '^'    '^' 
bisection  the  sum  of  two  sides  of  a  triangle  is  substituted  for  the  third 
side.    Moreover,  the  perimeter  can  never  pass  the  circumference,  since  a 
chord  is  always  less  than  its  arc. 

Now  let  AB  =  r  (?)  =  1  be  the  side  of  the  inscribed  hexagon.  Then 
by  the  formula  (418),  we  have 


=  1/2- 


CB  =  c  =  y2-V4-l  =  .51763809, 

which  is  therefore  the  side  of  a  regular  dodecagon.     Hence  the  perimeter 
of  the  dodecagtm  is 

.51763809  X  12  ^  6.21165708. 

Again,  let  the  side  of  the  inscribed  regular  polygon  of  24  sides  be  c\ 
and  we  have 


5'-|/s 


y^4  _  c«  =  y  2  -  ^4  -  (.51763809)«  =  .26105238; 


and  the  perimeter,        .26105238  x  24  =  6.26525722. 

Carrying  the  computation  forward  in  this  manner,  we  have  the  fol- 
lowing : 


APPLICATIONS    OF   DOCTRINE    OF    SIMILARITY. 


195 


C 

oc 

(N 

rH 

CO 

o 

T*< 

»-H 

o     ^ 

1 

ft 

O 

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^ 

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cc 

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l2 

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^^ 

o 

tH 

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cc 

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2> 

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tH 

r-i 

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00 

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C<i 

CO 

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CC 

ir- 

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tS 

o 

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CO 

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C<i 

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tj 

CO 

cc 

cc 

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5     "S 

03 

-tj' 

§ 

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o: 

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CO 

2> 

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*^    o 

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cc 

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c^ 

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cc 

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^^ 

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cc 

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00 

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•""• 

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2> 

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(N 

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y-< 

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1 

»c 

G<J 

tH 

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O 

o 

J"  t^ 

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r" 

o  -B 

• 

^*         <Q 

8     T^ 

• 

oj    C 

S    00 

H»» 

J3     O 

1" 

HM 

s  ^ 

1 — 1 

:^ 

4-       ^     X 

CO 

•-*» 

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,.».t^, 

-^    5 

<M          -  CO 

Hn 

N            -4» 

+ 

1 1 

1 — 1 

%  II 

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1 

s.     1 1 

1—^ 

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J    -^ 

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-      4- 

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4- 

ite  any  one  of  these  forms 
rresponding  polygon  is  3 
here  will  be  6  twos  ;  henc 

2-      2+       2- 

en  commencing  with  tl 
if  the  corresponding  po 

s 

1 

1 

r         vS- 

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^     £ 

►        c<? 

^ 

;c* 

r     1 

1 

^ 

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(^ 

J        c^ 

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►          (M 

*                 N^ 

G<? 

Mi 

•S 

1 

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J          «-y- 

L^ 

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1^ 

1 

C5 

1 

1 

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1 
1 

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— y- 
1 

-         4 

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^    1 — 1 

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des  of  th< 
bserve  th 

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;^. 

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►          «N 

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J         — y~ 

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7 

1 

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3             \ 

3        ">;:> 

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>            <?5 

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03     ^ 

196  ELEMENTARY    GEOMETRY, 

It  now  appears  that  the  first  four  decimal  figures  do  not  change  as 
the  number  of  sides  is  increased,  but  will  remain  the  same  7ww  far  soever 
we  2>Toceed.  When  the  foregoing  process  is  continued  till  5  decimals  be- 
come constant,  we  have  6. 28318 +  .  We  may  therefore  consider  6.28318 
as  app'oximatdy  the  circumference  of  a  circle  whose  radius  is  1, 

Hence,  letting  27r  stand  for  the  circumference,  we  have 
27r  =  6.28318 +  , 
and  TT  —  3.1416,  nearly,    q.  e.  d. 

421.  Scholium.— The  symbol  ;r  is  much  used  in  mathematics,  and 
signifies,  primarily,  the  aemi-drcumference  of  a  circle  whose  radius  is  1. 
^TT  is  therefore  a  symbol  for  a  quadrant,  90°,  or  a  right  angle.  \n  is 
equivalent  to  45",  and  2n  to  a  circumference,  the  radius  being  always 
supposed  1,  unless  statement  is  made  to  the  contrary.  The  numerical 
value  of  TT  has  been  sought  in  a  great  variety  of  ways,  all  of  which  agree 
in  the  conclusion  that  it  cannot  be  exactly  expressed  in  decimal  numbers, 
but  is  approximately  as  given  in  the  proposition.  From  the  time  of 
Archimedes  (287  B.C.)  to  the  present,  much  ingenious  labor  has  been 
bestowed  upon  this  problem.  The  most  expeditious  and  elegant  methods 
of  approximation  are  furnished  by  the  Calculus.  The  following  is  the 
value  of  n  extended  to  fifteen  places  of  decimals :  3.141592653589793. 


PROPOSITION    XII. 

422.    Theorem. — The  circumference  of  any  circle  is 
2rrr,  r  being  the  radius. 


'  Demonstration. 

The  circumferences  of  circles  being  to  each  other  as  their  radii  (417), 
and  27r  being  the  circumference  of  a  circle  whose  radius  is  1,  we  have 

27r    _  1^ 
circf.        r ' 

whence,  circf.  =  27rr.     Q.  b.  d. 

423.    Corollary. — The  circumference  of  any  circle  is 
ttD,  D  being  the  diameter. 


AFPLICATIONS    OF   DOCTRINE    OF    SIMILARITY.  197 


AREA    OF    THE    CIRCLE. 


PROPOSITION     XIII. 

424.    Theorem. — The  area  of  a  circle  whose  radius 
is  1,  is  TT. 

Demonstration. 

The  area  of  a  circle  is  ^r  x  circf.  (356).     When  r  =  1, 
circf.  =  2t(420); 
hence,  area  of  circle  whose  radius  is  1  =  ^  x  Stt  =  n^.    q.  e.  d. 


PROPOSITION    XIV. 

425.    Theorem. — The  area  of  any  circle  is  nr^,  r  being 
the  radias. 

Demonstration. 

The  areas  of  circles  being  to  each  other  as  the  squares  of  their  radii 
(413),  and  tt  being  the  area  of  a  circle  whose  radius  is  1,  we  have 

n  V 


area  of  any  circle       r' 
whence,  area  of  any  circle  =  tt? ■^     q.  e.  d. 

426.  Scholium  1. — Since  the  area  of  a  sector  is  to  the  area  of  the 
circle  of  the  same  radius  as  its  angle  is  to  4  right  angles  (359),  if  we 

d  TT  T^ 

represent  the  angle  of  the  sector  by  a°,  we  have  for  its  area  • 

427.  Scholium  2. — As  the  value  of  tt  cannot  be  exactly  expressed  in 
numbers,  it  follows  that  the  area  cannot.  Finding  the  area  of  a  circle  has 
long  been  known  as  the  problem  of  "  Squaring  the  Circle ;"  i.  e.,  find- 
ing a  square  equal  in  area  to  a  circle  of  given  radius.  Doubtless  many 
hare-brained  visionaries  or  ignoramuses  will  still  continue  the  chase  after 
the  phantom,  although  it  has  long  ago  been  demonstrated  that  the  diam- 


198  ELEMENTARY     GEOMETRY, 

eter  of  a  circle  and  its  circumference  are  incommensurable  by  any  finite 
unit.  It  is,  however,  an  easy  matter  to  conceive  a  square  of  the  same 
area  as  any  given  circle.  Thus,  let  there  be  a  rectangle  whose  base  is 
equal  to  the  circumference  of  the  circle,  and  whose  altitude  is  half  the 
radius ;  its  area  is  exactly  equal  to  the  area  of  the  circle.  Now,  let  there 
be  a  square  whose  side  is  a  mean  proportional  between  the  altitude  and 
base  of  this  rectangle ;  the  area  of  the  square  is  exactly  equal  to  the  area 
of  the  circle. 


PROPOSITION    XV. 

428.  Theorem.—//  a  -perpendicular  is  let  fall  from 
any  angle  of  a  triangle  upon  the  opposite  side  (or  on  the 
side  produced),  the  difference  of  the  squares  of  the  segments 
is  equivalent  to  the  difference  of  the  squares  of  the  other 
two  sides. 

Demonstration. 

Let  ABC  be  any  triangle,  and  CD  be  the  perpendicular  let  fall  from 
C  upon  AB  (or  AB  produced).  Call  the  sides  opposite  the  angles 
A,  B,  and  C,  a,  h,  and  c,  respectively;  and  let  the  segment  BD  =  m, 
AD  =  ti,  and  CD  =  p. 

Then  is    m'— n^  =  «'—&». 
For,  from  the  right-angled  tri- 
angle BCD, 

Also,  from  CDA, 

Whence,  o"  —  m^  =  &^  —  w 

or  m^  —  n^  =  a^  —  5' 

429.  Corollary, — Since 

m'  —  n*  =  (m  +  n)(m  —  n), 

and  a'-h'  =  {a-\-l){a-  I), 

m  +  n  (or  c)        a  —  b 

we  have  — r —  =   - — - 

a  +  0  m  —  n 


APPLICATIONS    OF   DOCTRINE    OF    SIMILARITY.  199 

430.  Scholium. — In  case  tlie  perpendicular  falls  without,  the  dis- 
tances BD  and  AD  are  still,  for  simplicity  of  expression,  spoken  of  as 
segments. 


431.  A  line  is  said  to  be  divided  in  Extreme  and 
Mean  Ratio  when  it  is  so  divided  that  the  whole  line  is  to  the 
greater  segment  as  the  greater  segment  is  to  tlie  less,  ?'.  e.y  when 
the  greater  segment  is  a  mean  proportional  between  the  whole 
line  and  the  less  segment. 


PROPOSITION    XVI. 

432.    Problem. — To  divide  a  line  in    extreme    and 
mean  ratio. 

Solution. 

Let  it  be  proposed  to  divide  the  line  AB  in  extreme  and  mean  ratio, 
ue,,  C  being  the  point  of  division,  so  that 

AB       AC 
AC  ~  CB" 

At  one  extremity  of  AB,  as  B,  erect  a 
perpendicular  BO,  and  make  it  equal  to 
iAB. 

From  0  as  a  centre,  with  OB  as  a  ra- 
dius, describe  a  circle. 

Draw  AO,  cutting  the  circumference 
in  D. 

Then  is  AD  the  greater  segment,  and  taking  AC  =  AD,  AB  is  divided 
in  extreme  and  mean  ratio  at  C. 

Demonstbation.of  Solution. 

Produce  AO  to  E. 

Now  ^  =  ^('n 

AB      AO  ^  ^' 

or,  by  inversion,  ^  =  ^. 

At       AB 


Fig.  214. 


200  ELEMENTARY    GEOME'tRT, 

By  division,  we  have 

AB  AD 


AE  -  AB  ~  AB  -  AD 

But,  as  DE  =  AB  (?), 

AE  -  AB  =  AE  -  DE  =  AD  =  AC; 

and  AB  -  AD  =  AB  -  AC  =  CB. 

A  Pi        Kf\ 

Hence,  substituting,  AC  ~  CB  '    ^  ^*  ^• 


PROPOSITION    XVII. 

433.  Problem. — To  inscribe  a  regular  decagon  in  a 
circle,  and  hence  a  regular  pentagon,  and  regular  polygons 
of  20,  40,  80,  etc.,  sides. 

Solution. 

Let  it  be  required  to  inscribe  a  regular  decagon  in  the  circle  whose 
centre  is  0  and  radius  OA. 

Divide  the  radius  OA  in 
extreme  and  mean  ratio,  as 
at  («). 

Then  is  ac,  the  greater 
segment,  the  side  of  the  in- 
scribed decagon,  ABODE, 
etc. 

To  prove  this,  draw  OA 
and  OB,  and  taking  CM  = 

°  Fig.  215. 

ac  =  AB,  draw  BM. 

^^^  OM  ~  MA '  ^y  construction ;  and,  as  CM  =  AB,  we  have 

0A_  AB 
AB  ~  MA  * 

Hence,  considering  the  antecedents  as  belonging  to  the  triangle  OAB, 
and  the  consequents  to  the  triangle  BAM,  we  observe  that  the  two  sides 
about  the  angle  A,  which  is  common  to  both  triangles,  are  proportional ; 
hence  the  triangles  are  similar  (373). 


EXERCISES.  201 

Therefore,  ABM  is  isosceles,  since  OAB  is,  and 
angle  BMA  =  A  =  OBA, 
and  MB  =  BA  =  OM. 

This  makes  0MB  also  isosceles,  and 

the  angle  0  =  OBM. 
Again,        the  exterior  angle  BMA  =  0  +  OBM  =  20; 
hence,  A  (which  equals  BMA)  =  20. 

Hence,  also,  OBA  (which  equals  A)  =  20. 

Wherefore,  0  is  |  the  sum  of  the  angles  of  the  triangle  OAB,  or  |  of 
2  right  angles,  =  y^  of  4  right  angles. 

The  arc  AB  is  therefore  the  measure  of  ^  of  4  right  angles,  and  is 
consequently  ^V  "^  t^e  circumference.  Hence  AB  is  the  chord  of  ^  of 
the  circumference,  and  if  applied,  as  AB,  BC,  CD,  DE,  etc.,  will  give  an 
equilateral  inscribed  decagon. 

Moreover  this  inscribed  polygon  is  equiangular,  and  hence  regular 
by  (272). 

To  construct  the  pentagon,  join  the  alternate  angles  of  the  decagon. 
To  construct  the  regular  polygon  of  20  sides,  bisect  the  arcs  subtended 
by  the  sides  of  the  decagon,  etc. 


MISCELLANEOUS    EXERCISES. 

434.  f.  Show  that  if  a  chord  of  a  circle  is  conceived  to  re- 
volve, varying  in  length  as  it  revolves,  so  as  to  keep  its  extremities 
in  the  circumference  while  it  constantly  passes  through  a  fixed 
point,  the  rectangle  of  its  segments  remains  constant. 

2.  The  two  segments  of  a  chord  intersected  by  another  chord 
are  6  and  4,  and  one  segment  of  the  other  chord  is  3.  What  is 
the  other  segment  of  the  latter  chord  ? 

3.  Show  how  Propositions  I,  II,  and  III  may  be  considered  as 
different  cases  of  one  and  the  same  proposition. 

Suggestions. — By  stating  Propositions  I  and  IT  thus,  The  distances 
from  the  irUer section  of  the  linen  to  their  intersections  with  tlie  circumference^ 
what  follows?  In  Fig.  204,  if  the  secant  AO  becomes  a  tangent,  what 
does  OD  become  ? 


202  ELEMENTARY    GEOMETRY. 

4.  In  a  triangle  whose  sides  are  48,  36,  and  50,  where  do  the 
bisectors  of  the  angles  intersect  the  sides  ? 

5.  In  the  last  example,  find  the  lengths  of  the  bisectors. 

6.  A  and  B  have  farms  of  similar  shape,  with  their  homolo- 
gous sides  on  the  same  road.  A's  is  150  rods  on  the  road,  and 
B's  200  rods.     How  does  A's  farm  compare  with  B's  in  size  ? 

7.  Draw  two  similar  triangles  with  their  homologous  sides  in 
the  ratio  of  3  to  5,  and  divide  them  into  equal  partial  triangles, 
showing  that  their  areas  are  as  3^  to  5^,  that  is,  as  9  to  25. 

8.  What  are  the  relative  capacities  of  a  5-inch  and  a  7-inch 
stove-pipe  ? 

9.  If  a  circle  whose  radius  is  34  is  divided  into  5  equal  parts 
by  concentric  circumferences,  what  are  the  diameters  of  the  sev- 
eral circles  ? 

Solve  geometrically  as  well  as  numerically. 

10.  The  projection  of  one  line  upon  another  in  the  same  plane 
is  the  distance  between  the  feet  of  two  perpendiculars  let  fall 
from  the  extremities  of  the  former  upon  the  latter.  Show  that 
this  projection  is  equal  to  the  square  root  of  the  difference  be- 
tween the  square  of  the  line  and  the  square  of  the  difference  of 
the  perpendiculars. 

11.  The  three  sides  of  a  triangle  being  4,  5,  and  6,  find  the 
segments  of  the  last  side  made  by  a  perpendicular  from  the  op- 
posite angle.  Ans.  3.75  and  2.25. 

12.  Same  as  above,  when  the  sides  are  10,  4,  and  7,  and  the 
perpendicular  is  let  fall  from  the  angle  included  by  the  sides  10 
and  4.     Draw  the  figure.     Why  is  one  of  the  segments  negative  ? 

13.  What  is  the  area  of  a  regular  octagon  inscribed  in  a  circle 
whose  radius  is  1  ?  What  is  its  perimeter?  What  if  the  radius 
is  10? 

14.  What  is  the  side  of  an  equilateral  triangle  inscribed  in  a 
circle  whose  radius  is  1  ? 


EXERCISES.  203 

15.  What  is  the  side  of  a  regular  inscribed  decagon  in  a  circle 
whose  radius  is  4 ?  What  the  side  of  the  inscribed  pentagon? 
What  is  the  area  of  each  ? 

16.  Draw  two  squares,  and  construct  two  others,  one  equal  to 
their  sum,  and  the  other  to  their  difference. 

17.  Draw  any  two  polygons,  and  construct  two  squares,  one 
equivalent  to  their  sum,  and  the  other  equivalent  to  their  differ- 
ence. 

18.  Show  that  the  length  of  a  degree  in  any  circle  is  -^y 

and  hence  that  the  lengths  of  degrees  in  different  circles  are  to 
each  other  as  the  radii  of  the  circles. 

19.  What  is  the  length  of  a  minute  on  a  circle  whose  radius 
is  10  miles  ? 

20.  Calling  the  equatorial  radius  of  the  earth  3962.8  miles, 
what  is  the  length  of  a  degree  on  the  equator  ? 

21.  How  many  degrees  in  the  arc  of  a  circle  which  is  equal  in 
length  to  the  radius  ? 

22.  Compute  the  area  of  the  triangle  whose  sides  are  20,  30, 
and  40. 

Find  the  segments  of  the  base  (40)  by  (428)-  Hence  the  perpendic- 
ular. 

23.  Given  the  side  of  a  regular  inscribed  pentagon,  as  16,  to 
find  the  side  of  the  similar  circumscribed  polygon. 

24.  Prove  that  if  a  triangle  is  circumscribed  about  a  given 
triangle  by  drawing  lines  through  the  vertices  of  the  given  tri- 
angle and  parallel  to  the  opposite  sides,  the  area  of  the  circum- 
scribed triangle  is  four  times  that  of  the  given  triangle. 

25.  Prove  that  the  bisectors  of  the  angles  of  a  triangle  pass 
through  a  common  point. 

26.  Prove  that  the  perpendiculars  to  the  three  sides  of  a  tri- 
angle at  their  middle  points  pass  through  a  common  point. 


304: 


ELEMENTA  R  Y     GEOMETR  Y. 


27.  The  three  perpendiculars  drawn  from  the  angles  of  a  tri- 
angle upon  the  opposite  sides  intersect  in  a  common  point. 

Draw  through  the  vertices  of  the 
triangle  lines  parallel  to  the  opposite 
sides.  The  proposition  may  then  be 
brought  under  the  preceding. 

28.  The  following  triangles  are 
similar — viz.,  BOE,  BDC,  AOD,  and 
AEC,  each  to  each  ;  also  EOF,  BDA, 
DOC,  and  CFA.     Prove  it. 

Fig.  216. 


435.  The  Medial  Lines  in  a  triangle  are  the  lines  drawn 
from  the  vertices  to  the  middle  points  of  the  opposite  sides. 

29.  The  three  medial  lines  of  a  triangle  mutually  trisect  each 
other,  and  hence  intersect  in  a  common  point. 

To  prove  that  OE  =  ^BE  (Fig.  317),  draw  FC  parallel  to  AD  until  it 

meets  BE  produced.    Then  the  triangles  AEO  and  FEC  are  equal  (?); 

whence 

EF  =  OE. 

Also,  BO  =  OF  (?). 

Having  shown  that 

OE  =  iBE, 
by  a  similar  construction  we  can  show  that 
OD  =  iAD 
Finally,  we  may  show  that  the  medial  line 
from  C  to  AB  cuts  off  ^  of  BE,  and  hence  cuts  BE 
at  the  same  point  as  does  AD. 


I- — ^c 


Fig.  217. 


Another  Demonstration. — Lines  through  0  parallel  to  the  sides 
trisect  the  sides,  etc. 

Still  Another.— Without  EF  and  FC,  draw  ED,  and  prove  by  simi- 
lar triangles. 


4-  CHAPtfeR  n; 

SOLID       GEOMETRY.* 


OF    STRAIGHT     LINES    AND     PLANES. 

436.  Solid  Geometry  is  that  department  of  Geometry 
in  which  the  magnitudes  treated  are  not  limited  to  a  single 
plane.    "  ^ 

437.  A  Plane  (or  a  Plane  Surface)  is  a  surface  such 
that  a  straight  line  joining  any  two  points  in  it  lies  wholly  in  the 
surface. 


PLANE,     HOW     DETERMINED. 

438.  A  plane  is  said  to  be  Determined  by  given  conditions 
which  fix  its  position. 

All  planes  are  considered  as  indefinite  in  extent,  unless  the 
contrary  is  stated. 


*  In  some  respects,  perhaps,  " Geometry  of  Space"  is  preferable  to  this 
term  ;  but,  as  neither  is  free  from  objections,  and  as  this  has  the  advantage 
of  simplicity  and  long  use,  the  author  prefers  to  retain  it. 


206  ELEMENTARY    GEOMETRY, 


PROPOSITION    I. 

439.  Theorem. — Three  points  not  in  the  same  straight 
line  determine  a  plane. 

Demonstration. 

Let  A,  B,  and  C  be  three  points  not  in  the  same  straight  line. 

Then  one  plane  can  be  passed  through 
them,  and  only  one;  i.  c,  they  determine  the 
position  of  a  plane. 

For,  pass  a  straight  line  through  any  two  of 
these  points,  as  A  and  B.  Now,  conceive  any 
plane  containing  these  two  points ;  then  will  the 
line  passing  through  them  lie  wholly  in  the 
plane  (437).     Conceive  this  plane  to  revolve  on  ^'9-  2i8. 

the  line  as  an  axis  until  the  point  C  falls  in  the  plane.     Thus  we  have  one 
plane  passed  through  the  three  points. 

That  there  can  be  only  one  is  evident,  since  when  C  falls  in  the  plane, 
if  the  plane  be  revolved  either  way,  C  will  not  be  in  it.  The  same  may 
be  shown  by  first  passing  a  plane  through  B  and  C,  or  A  and  C.  There 
is,  therefore,  only  one  position  ot  the  plane  in  which  it  will  contain  the 
third  point,    q.  e.  d. 

440.  Corollary  1. — A  line  and  a  point  witJwut  it  de^ 
termine  a  plane.  ^ 

441.  Corollary  2. — Through  one  line,  or  two  points,  an 
infinite  number  of  planes  can  he  passed.     , 

442.  Corollary  3. — The  intersection  of  two  planes  is  a 
straight  line. 

For  two  planes  cannot  have  even  three  points,  not  in  the  same  straight 
line,  common,  much  less  an  indefinite  number,  which  would  be  required 
if  we  conceived  the  intersection  (that  is,  the  common  points)  to  be  in  any 
other  than  a  straight  line. 


443.  The  Trace  of  one  plane  in  another  is  their  intersection. 


STRAIGHT   LINES    AND    PLANES.  207 


PROPOSITION    II. 

444.  Theorem. — Two  intersecting  lines  determine  the 
position  of  a  plane. 

Demonstration. 

For,  the  point  of  intersection  may  be  taken  as  one  of  the  three  points 
requisite  to  determine  the  position  of  a  plane,  and  any  two  other  points, 
one  in  each  of  the  lines,  as  the  other  two  requisite  points.  Now,  the 
plane  passing  through  these,  points  contains  lx)th  the  lines,  for  it  contains 
two  points  in  each.  q.  e.  d. 


PROPOSITION    III. 

445.  Theorem. — Two  parallel  lines  determine  the  po- 
sition of  a  plane. 

Demonstration. 

For,  pass  a  plane  through  one  of  the  parallels,  and  conceive  it  revolved 
until  it  contains  some  point  of  the  second  parallel.  Now,  if  the  plane  be 
revolved  either  way  from  this  position,  the  point  will  be  left  without  it. 
Hence,  it  is  the  only  plane  containing  the  first  parallel  and  this  point  in 
the  second. 

But  parallels  lie  in  the  same  plane  (120, 121),  whence  the  plane  of  the 
parallels  must  contain  the  first  line  and  the  specified  point  in  the  second. 

Therefore,  the  plane  containing  the  first  line  and  a  point  in  the  second 
is  the  plane  of  the  parallels,  and  is  fixed  in  position,     q.  e.  d. 

446.  Scholium. — When  a  plane  is  determined  by  two  lines,  accord- 
ing to  either  of  the  last  two  propositions,  it  is  spoken  of  as  the  Plane  of 
the  Lines.    In  like  manner,  we  may  speak  of  the  Plane  of  Three  Points. 


RELATIVE   POSITION   OF  A   LINE  AND 
A    PLANE. 

447.    A  line  may  have  one  of  three  positions  in  relation  to  a 
plane :  {a)  It  may  be  perpendicular,  (b)  oblique,  or  (c)  parallel 


208 


ELEMENTARY    GEOMETRY. 


OF  LINES  PERPENDICULAR  TO  A  PLANE. 

448.  A  line  is  said  to  Pierce  a  plane  at  the  point  where  it 
passes  through  it. 

449.  The  point  where  a  perpendicular  meets,  or  pierces,  a 
plane  is  called  its  Foot. 

450.  A  Perpendicular  to  a  Plane  is  a  line  which  is 
perpendicular  to  all  lines  of  the  plane  passing  through  its  foot, 
and  hence  to  every  line  of  the  plane.  Conversely,  the  plane  is 
perpendicular  to  the  line. 

451.  The  Distance  of  a  point  from  a  plane  is  the  length 
of  the  perpendicular  let  fall  from  the  point  upon  the  plane. 


PROPOSITION    IV. 

452.  Theorem. — A  line  which  is  perpendicular  to  two 
lines  of  a  plane,  at  their  intersection,  is  perpendicular  to 
the  plane. 

Demonstration^. 

Let  PD  be  perpendicular  to  AB  and  CF  at  D. 

Then  is  it  perpendicular  to  MN,  the 
plane  of  the  lines  AB  and  CF. 

Let  OQ  be  any  other  line  of  the 
plane  MN,  passing  through  O.  Draw 
FB  iutersecting  the  three  lines  AB,  CF, 
and  OQ  in  B,  E,  and  F.  Produce  PD 
to  P',  making  P'D  =  PD,  and  draw  PF, 
PE,  PB,  P'F,  PE,  P'B. 

Then  is      PF  ^  P'F, 
and  PB  =  P'B, 

since  FD  and  BD  are  perpendicular  to 
PP',  and 

PD  =  P'D(96) 


Fig.  219. 


STRAIGHT    LINES    AND    PLANES,  209 

Hence,  the  triangles  PFB  and  PTB  are  equal  (305) ;  and  if  PFB  be 
revolved  upon  FB  till  P  falls  at  P',  PE  will  fall  in  P'E. 

Therefore,  OQ  has  E  equally  distant  from  P  and  P',  and  as  D  is  also 
equidistant  from  the  same  points,  OQ  is  perpendicular  to  PD  at  D  (98). 

Now,  as  OQ  is  any  line,  PD  is  perpendicular  to  any  line  of  the  plane 
passing  through  its  foot,  and  consequently  perpendicular  to  the  plane 
(460).    Q.  E.  D. 

463.  Corollary. — If  one  of  two  perpendiculars  revolves 
about  the  other  as  an  axis,  its  path  is  avlane  perpendicu- 
lar to  the  axis,  and  this  plane  contains  all  the  perpendicu- 
lars to  the  axis  at  the  common  point. 

ThUvS,  if  AB  revolves  about  PP'  as  an  axis,  it  describes  the  plane  MN, 
and  MN  contains  all  the  perpendiculars  to  PP'  at  D.  For,  if  there  could 
be  a  perpendicular  to  PP  at  D  which  did  not  lie  in  the  plane  M  N,  there 
would  be  two  perpendiculars  to  PP'  at  D,  both'^' lying  in  the  same  plane, 
which  is  impossible  (88).  '  ^ 


PROPOSITION    V. 

454.   Theorem. — M  any  point  in  a  plane  one  perpen- 
dicular can  he  erected  to  the  plane,  and  only  one. 

Demonstration. 

Let  it  be  required  to  show  that  one  perpendicular,  and  only  one,  can 
be  erected  to  the  plane  MN  at  D. 

Through  D  draw  two  lines. of  the  plane,  as 
AB  and  CE,  at  right  angles  to  each  other.  CE 
being  perpendicular  to  AB,  let  a  line  be  con- 
ceived as  starting  from  the  position  ED  to  re- 
volve about  A  B  as  an  axis.  It  will  remain  per- 
pendicular to  AB  (453).  Conceive  it  to  have 
passed  to  P'D.  Now,  as  it  continues  to  revolve, 
P'DC  diminishes  continuously,  and  at  the  same 

,  •  .  .  rig.  220. 

rate  as  P'DE  increases;   hence,  in  one  position 

of  the  revolving  line,  and  in  only  one,  as  PD,  PDE  =  PDC,  and  PD  is 

perpendicular  to  CE  (86). 

Again,  any  line  which  is  perpendicular  to  MN  at  D  is  perpendicular 


210 


ELEMENTA  R  Y     GEOMETR  Y. 


to  AB  and  CE  (450).  But  the  plane  of  the  lines  PD  and  DE  contains  all 
lines  perpendicular  to  AB  at  D.  Hence,  PD  is  perpendicular  to  the  plane 
(452),  and  is  the  only  perpendicular,     q.  e.  d. 


PROPOSITION    VI. 

455.   Theorem — From  a  point  without  a  plane  one 
perpendicular  can  he  drawn  to  the  plane,  and  only  one. 

4 

Demonstration. 


Let  It  be  required  to  show  that  one  perpendicular  can  be  drawn  from 
P  to  the  plane  MN,  and  only  one. 

Take  RS  as  an  aux- 
iliary plane,  and  at  any 
point  as  C  erect  DC  per- 
pendicular to  RS. 

Now  place  the  plane 
RS  in  coincidence  with 
MN,  and  move  it  in  MN 
till  the  perpendicular  DC 
passes  through  P. 

Then  DC,  which  passes  Fig.  221 

through  P  and  is  pei-pen- 
dicular  to  RS,  is   perpendicular  to   MN,  with   which   RS  is  coincident. 

Q.   E.   D. 

To  prove  that  there  can  be  but  one  perpendicular  from  P  to  MN,  sup- 
pose that  there  could  be  two,  as  PA  and  PF. 

Draw  FA. 

Then  since  FA  is  a  line  of  the  plane,  and  PF  and  PA  are  perpendic- 
ulars to  the  plane,  PFA  and  PAF  are  both  right  angles  (?),  and  the  tri- 
angle PFA  has  two  right  angles,  which  is  absurd.  Hence  there  can  be 
but  one  perpendicular  from  P  to  MN.     Q.  e.  d. 

456.  Corollary.  —  The  perpendicular  is  the  shortest  line 
that  can  he  drawn  to  a  plane  from  a  point  without. 

Thus,  let  PA  be  a  perpendicular  and  PF  any  oblique  line, 
PA  <  PF  (?). 


STRAIGHT   LINES    AND    PLANES, 


211 


PROPOSITION    VII. 

457.  Theorem. — Conversely  to  the  last.  Through  a  given 
point  in  a  line,  one  plane  can  be  passed  perpendicular  to 
the  line,  and  only  one. 

Demonstration. 

Let  D  be  the  point  in  the  line  PG. 

Pass  two  lines  through  D,  as 
EF  and  AB,  each  perpendicular  to 
PD ;  the  plane  of  these  lines  is  per- 
pendicular to  PD.     Q.  E.  D. 

To  show  that  but  one  plane  can 
be  passed  through  D  perpendicular 
to  PG,  assupie  that  M'N'  is  another 
plane  passing  through  D,  and  per- 
pendicular to  PG,  but  not  contain- 
ing BD.  Through  PD  and  BD  pass 
a  plane,  and  let  BD  be  its  intersec- 
tion with  M'N'.  Then,  on  the 
hypothesis  that  M'N'  is  perpendic- 
ular to  PG,  B'DP  is  a  right  angle,  and  we  have  two  lines  in  the  same 
plane  with  PG,  and  perpendicular  to  it  at  the  same  point,  which  is 
absurd.  Hence  there  can  be  but  one  plane  perpendicular  to  PG  and  pass- 
ing through  D.     Q.  e.  d. 


PROPOSITION    VIII. 

458.  Theorem. — //  from  the  foot  of  a  perpendicular 
to  a  plane  a  line  is  drawn  at  right  angles  to  any  line  of 
the  plane,  and  their  intersection  is  joined  with  any  point 
in  the  perpendicular,  the  last  line  is  perpendicular  to  the 
line  of  the  plane. 

Demonstration. 

From  the  foot  of  the  perpendicular  PD  (Fig.  223)  let  DE  be  drawn 
perpendicular  to  AB,  any  line  of  the  plane  MN,  and  E  joined  with  0,  any 
point  of  the  perpendicular. 


212 


ELEMENT  A  RY    G  E03IETR  Y, 


Then  is  OE  perpendicular  to  AB. 

Take  EF  =  EC,  and  draw  CD,  FD,  CO,  and 
FO.    Now, 

CD  =  DF  (?), 

whence  CO  =  FO  (?), 

and  OE  has  0  equally  distant  from  F  and  C,  and 
also  E.    Therefore,  OE  is  perpendicular  to  AB  (?). 

Q.  E.  D. 


Fig.  223. 


459.  Corollary. — The  line  DE  measures  the  shortest 
distance  between  PD  and  AB. 

For  a  line  drawn  from  E  to  any  other  point  in  PD  than  D,  as  Ea,  is 
longer  than  DE  (?). 

Again,  if  from  any  other  point  in  AB,  as  C,  a  line  be  drawn  to  D,  it  is 
longer  thari  DE  (?) ;  and  if  drawn  from  C  to  «,  any  other  point  in  PD 
than  D,  Oa  is  longer  than  CD  (?),  and  consequently  longer  than  DE  (?). 


PROPOSITION    IX. 

460.   Theorem. — //  one  of  two  parallels  is  perpendic- 
ular to  a  plane,  the  other  is  perpendicular  also. 


Demonstration. 
Let  AB  be  parallel  to  CD  and  perpendicular  to  the  plane  MN. 

Then  is  CD  perpendicular  to  MN. 

For,  drawing  BD  in  the  plane  MN,  it  is  pcr- 
pendiculaf  to  AB  (?),  and  consequently  to  CD  (?). 
Through  D  draw  EF  in  the  plane  and  perpendic- 
ular to  BD,  and  join  D  with  any  point  in  AB,  as 
A;  then  is  EF  perpendicular  to  AD  (?). 

Now,  EF  being  perpendicular  to  two  lines,  AD 
and  BD,  of  the  plane  ABDC,  is  perpendicular  to  F'g-  224. 

the  plane,  and  hence  to  any  line  of  the  plane  passing  through  D,  as  CD. 

Therefore,  CD  is  perpendicular  to  BD  and  EF,  and  consequently  to  the 
plane  MN  (?).     Q.  e.  d. 


STRAIGHT    LINES   /iND    PLANES.  213 

461.  Corollary.— T/fo  lines  which  are  perpendicular 
to  the  same  plane  are  parallel. 

Thus,  AB  and  CD  being  perpendicular  to  the  plane  MN  are  parallel. 
For,  if  AB  is  not  parallel  to  CD,  draw  a  line  through  B  which  shall  be. 
By  the  Proposition,  this  line  is  perpendicular  to  MN,  and  hence  must 
coincide  with  AB  (454)- 


PROPOSITION    X. 

462.    Theorem. —Ta^o  lines  parallel  to  a  third  not  in 
their  own  plane  are  parallel  to  each  other. 


D  EMO  N  STR  ATION-V 

Let  AB  and  CD  be  parallel  to  EF. 

Then  are  they  parallel  to  each  other. 

For,  through  F',  any  point  in  EF,  pass  a  plane 
MN  perpendicular  to  EF. 

Now  AB  and  CD  are  respectively  perpendicu- 
lar to  MN  (?),  and  hence  are  parallel  to  each  other 
(?).     Q.  E.  D. 


Fig.  225. 


OF     LINES     OBLIQUE     TO     A     PLANE. 

463.  An  Oblique  Line  is  a  line  which  pierces  the  plane 
(if  sufficiently  produced),  but  is  not  perpendicular  to  the  plane. 

464.  The  Projection  of  a  Point  on  a  plane  is  the  foot 
of  the  perpendicular  from  the  point  to  the  plane. 

465.  The  Projection  of  a  Line  upon  a  plane  is  the 
locus  of  the  projection  of  the  point  which  generates  the  line. 


214 


ELEMENTARY    GEOMETRY, 


PROPOSITION    XI. 

466.  Theorem. — The  projection  of  a  straight  line  upon 
a  plane  is  a  straight  line. 


Demonstration. 
Let  AB  be  any  line  and  MN  the  plane  upon  which  it  is  projected. 

Then  is  the  projection  of  AB  in  MN 
a  straight  line. 

Let  P  be  a  point  in  AB,  and  D  its 
projection  in  IMN. 

Pass  a  plane,  S,  through  AB  and  PD 
(444),  and  let  CE  be  its  trace  in  MN. 

Now  let  P'  be  any  point  in  AB 
other  than  P,  and  let  D'  be  its  projec- 
tion in  MN. 

As  PD  and  P'D'  are  perpendicular 
to  MN,  they  are  parallel  to  each  other 
(461),  and  a  plane  may  be  passed  throug'i  them  (445). 

But  the  plane  of  PD  and  P'D'  is  S,  since  it  contains  PD  and  P'  (440). 

Therefore  D'  lies  in  S,  and  as  it  lies  in  MN,  it  is  in  the  trace  of  S  in 
MN,  which  trace  is  a  straight  line  (442). 

Hence,  as  P'  is  any  point  in  AB,  the  projection  of  every  point  of  AB  is 
in  a  straight  line.     q.  e.  d. 


•■■■^■^BfS^H 


Fig.  226. 


467.  Corollary. — The  projection  of  a  line  upon  a  plane 
is  the  trace  of  a  plane  containing  the  line  and  the  projec- 
tion of  any  point  of  the  line. 


468.  The  Projecting  Plane  is  the  plane  of  a  line  and  its 
projection  upon  another  plane. 

469.  The  Plane  of  Projection  is  the  plane  upon  which 
a  point  or  a  line  is  projected. 

470.  The  Inclination  of  a  Line  to  a  plane  is  the  angle 
included  between  the  line  and  its  projection. 


STRAIGHT    LINES    AND    PLANES, 


215 


PROPOSITION    XII. 

471.  Theorem. — If  from  any  point  in  a  perpendicular 
to  a  plane,  oblique  lines  are  drawn  to  the  plane,  those  which 
pierce  the  plane  at  equal  distances  from  the  foot  of  the 
perpendicular  are  equal ;  and  of  those  which  pierce  the 
plane  at  unequal  distances  from  the  foot  of  the  perpendic- 
ular, those  which  pierce  at  the  greater  distances  are  the 
greater. 

Demonstration. 

Let  PD  be  a  perpendiculdp  to  the  plane  MN,  and  PE,  PE',  PE",and 
PE' "  be  oblique  lines  piercing  the  plane  at  equal  distances  ED,  ED,  E"D, 
and  E  "D  from  the  foot  of  the  perpendicular. 

Then  PE  =  PE'  =  PE"  =  PE". 

For  each  of  the  triangles  PDE,  PDE',  etc., 
has  two  sides  and  the  included  angle  equal 
to  the  corresponding  parts  in  the  other. 

Again,  let  FD  be  longer  than  E'D. 

Then  is  PF  >  PE'. 

For,  take  ED  =  E'D ;  then  PE  =  PE',  by 
the  preceding  part  of  the  demonstration. 
But  PF  >  PE,  by  (113).     Hence,  PF  >  PE'.    Q.  e.  d. 


472.  Corollary  1. — The  angles  which  oblique  lines 
drawn  from  a  common  point  in  a  perpendicular  to  a 
plane,  and  piercing  the  plane  at  equal  distances  from  the 
foot  of  the  perpendicular,  make  with  the  perpendicular, 
are  equal ;  and  the  inclinations  of  such  lines  to  the  plane 
are  equal. 

Thus,  the  equality  of  the  triangles,  as  shown  in  the  demonstration, 
shows  that 

EPD  =  E'PD  =  E'PD  =  E  "PD, 


and 


PED  =  PE'D  =  PE"D  =  PE  "D. 


473.    Corollary  2. — Conversely,    //   the    angles   which 
oblique  lines  drawn  from  a  point  in  a  perpendicular  to  a 


216  ELEMENTARY    GEOMETRY. 

plane,  make  with  the  perpendicular,  are  equal,  the  lines 
are  equal,  and  pierce  the  plane  at  equal  distances  from 
the  foot  of  the  perpendicular. 

Thus,  let  E'PD  =  E"PD  ; 

then  the  right-angled  triangles  PDE'  and  PDE"  are  equal  (?).     Hence, 
PE'  =  PE",  and  DE'  =  DE". 

474.  Corollary  3. — Lines  drawn  from  the  same  point 
in  a  perpendicular,  and  equally  inclined  to  the  plane,  are 
equal,  and  pierce  the  plane  at  equal  distances  from  the 
foot  of  the  perpendicular. 

475.  Corollary  4. — Equal  oblique  lines  from  the  same 
point  in  the  perpendicular,  pierce  the  plane  at  equal  dis- 
tances fror)v  the  foot  of  the  perpendicular,  a>re  equally 
inclined  to  the  plane,  and  also  to  the  perpendicular. 

Since  the  right-angled  triangles  PDE'  and  PDE"  have  their  altitudes 
and  hypotenuses  equal,  the  triangles  are  equal  (309),  and 

DE'  =  DE",  PE'D  =  PE "D,  and  E'PD  ==  E  "PD. 


OF     LINES     PARALLEL     TO     A     PLANE. 

476.  A  Line  is  Parallel  to  a  Plane  when  it  is  paral- 
lel to  its  projection  in  that  plane. 


PROPOSITION     XIII. 

477.  Theorem. — A  line  parallel  to  a  plane  is  every- 
where equidistant  from  the  plane,  and  hence  can  never 
meet  the  plane ;  and,  conversely,  a  straight  line  which  can- 
not meet  a  plane  is  parallel  to  it. 

Demonstration. 

The  distance  between  a  point  in  the  line  and  the  plane  being  the  per- 
pendicular (451),  is  also  the  distance  between  the  point  and  the  projec- 


STRAIGHT   LINES    AND    PLANES.  217 

tiori  of  the  line  (464).  But  this  is  everywhere  the  same  (476,  136)- 
Hence  a  line  parallel  to  a  plane  is  everywhere  equidistant  from  it,  and 
therefore  can  never  meet  it.     Q.  e.  d. 

Conversely ;  A  line  which  meets  a  plane  meets  it  in  the  projection  of 
the  line  in  the  plane,  since  the  projecting  plane  contains  all  the  per- 
pendiculars, or  shortest  lines,  from  the  line  to  the  plane.  Hence  a  line 
which  never  meets  a  plane  is  parallel  to  its  projection  in  that  plane,  that 
is,  to  the  plane  itself  (476).    Q-  e.  d. 


PROPOSITION    XIV. 

478.   Theorem. — Either  of  two  parallel  lines  is  paral- 
lel to  every  plane  containing  the  other. 

Demonstration. 

Let  AB  and  CD  be  two  parallel  lines,  and  MN  a  plane  containing  CD. 

Then  is  AB  parallel  to  the  plane  MN. 

Since  AB  and  CD  are  in  the  same  plane 
(?),  and  as  the  intersection  of  their  plane 
with  MN  is  CD  (?),  if  AB  meets  the  plane 
MN,  it  must  meet  it  in  CD,  or  CD  produced. 
But  this  is  impossible  (?). 

Whence  AB  is  parallel  to  MN  (477). 

Q.  E.  D.  Fig-  228. 

479.  Corollary  1.  A  line  which  is  parallel  to  a  line  of 
a  plane  is  parallel  to  the  plane. 

480.  Corollary  2. — Through  any  given  line,  a  plane 
may  he  passed  parallel  to  any  other  given  line  not  in  the 
plane  of  the  first. 

For,  through  any  point  of  the  line  through  which  the  plane  is  to  pass, 
conceive  a  line  parallel  to  the  second  given  line.  The  plane  of  the  two 
intersecting  lines  is  parallel  to  the  second  given  line  (?). 

481.  Corollary  3. — Through  any  point  in  space  a  plane 
may  he  passed  parallel  to  any  two  lines  in  space. 

For,  through  the  given  point  conceive  two  lines  respectively  parallel 
to  the  given  lines ;  then  is  the  plane  of  these  intersecting  lines  parallel  to 
the  two  given  lines  (?). 
10 


218 


ELEMENTARY    GEOMETRY. 


PROPOSITION    XV. 

482.  Theorem. — Of  two  lines  perpendicular  to  each 
other,  if  one  is  perpendicular  to  a  plane  the  other  is  par- 
allel to  the  plane. 

Demonstration. 

Let  AB  and  PD  be  perpendicular  to  each  other,  and  PD  perpendicu- 
lar to  the  plane  MN. 

Then  is  AB  parallel  to  MN. 

If  AB  does  not  intersect  PD, 
through  any  point  in  PD,  as  G, 
draw  A'B'  parallel  to  AB ;  then  is 
it  perpendicular  to  PD  (32,  foot- 
note). 

Let  CE  be  the  projection  of 
A'B'  in  the  plane  MN.  Then  is  H 
the  point  where  PD  pierces  the 
plane  in  CE  (?). 

Hence  A'B'  is  parallel  to  its 
projection  CE  (?),  and  consequently 
parallel  to  the  plane  MN. 

Therefore  AB  is  parallel  to  CE  (?),  and  consequently  to  the  plane  MN 
(479).    Q.  E.  D. 

483.  Corollary. — A  line  and  a  plane  which  are  both 
perpendicular  to  the  same  line  are  parallel. 


Fia.  229. 


RELATIVE  POSITION  OF  TWO  PLANES. 


OF     PARALLEL     PLANES. 

484.  Parallel  Planes  are  such  that  either  is  parallel  to 
any  line  of  the  other. 

485.  The  Distance  between  Two  Parallel  Planes 

at  any  point  is  measured  by  the  perpendicular. 


STRAIGHT    LINES    AND    PLANES. 


219 


PROPOSITION    XVI. 

486.  Theorem. — Parallel  planes  are  everywhere  equi- 
distant and  hence  can  never  meet. 


Demonstration. 


Let  P  and  Q  be  two  parallel 
planes. 

Then  are  they  everywhere  equi- 
distant, and  hence  can  never  meet. 

Let  A  and  B  be  any  two  points 
in  P,  and  pass  a  line  through  them. 

Since  Q  is  parallel  to  P,  it  is 
parallel  to  the  line  AB  (484).  And 
since  it  is  parallel  to  A B  it  is  every- 
where equidistant  from  AB. 

Hence  A  and  B,  any  two  points 
in  P,  are  equidistant  from  Q,  and 
consequently  P  and  Q  can  never 
meet.    q.  e.  d. 


Fig.  230. 


PROPOSITION    XVII. 

487.    Theorem. — Two    planes   perpendicular  to   the 
same  line  are  parallel  to  eoA^h  other. 

Demonstration. 

Let  P  and  Q  be  two  planes  perpen- 
dicular to  the  line  AB. 

Then  are  P  and  Q  parallel. 

For  any  line  in  one  plane  is  parallel 
to  its  projection  in  the  other,  since  any 
line  in  either  plane  is  perpendicular  to 
AB  (?). 

Hence  either  plane  is  parallel  to  any 
line  of  the  other  (476),  and  therefore  the 
planes  are  parallel  to  each  other,    q.  e.  d.  Fig.  231. 


^ 


220 


ELEMENTARY     GEOMETRY. 


PROPOSITION    XVIII. 

488.  Theorem.— 7/  a  plane  intersects  two  parallel 
planes,  the  lines  of  intersection  are  parallel. 

Demonstration. 

Let  RS  intersect  the  parallel  planes  MN  and  PQ  in  AB  and  CD. 

Then  is  AB  parallel  to  CD. 

For,  if  AB  and  CD  could  meet,  the  planes 
MN  and  PQ  would  meet,  as  every  point  in  A B  is 
in  MN,  and  every  point  in  CD  in  PQ.  Hence, 
AB  and  CD  lie  in  the  same  plane,  and  do  not 
meet  how  far  soever  they  be  produced  (132) ; 
they  are  therefore  parallel,    q.  e.  d. 

489.  Corollary. — Parallel  lines  in- 
tercepted between  parallel  planes  are  equal.    Fig.  232. 

Thus,  AC  =  BD,  if  they  are  parallel.  For,  the  intersections  AB  and 
CD,  of  the  plane  of  these  parallels,  are  parallel  (?),  and  the  figure  ABDC 
is  a  parallelogram ;  whence,  AC  =  BD  (?). 


PROPOSITION    XIX. 

490.    Theorem.— t^  line  which  is  perpendicular  to 
one  of  two  parallel  planes,  is  perpendicular  to  the  other 

also. 

Demonstration. 

Let  MN  and  PQ  be  two  parallel  planes;  and 
let  AB  be  perpendicular  to  PQ. 

Then  is  AB  perpendicular  to  MN. 

For,  pass  any  plane  through  AB,  and  let  AC 
and  BD  be  its  intersections  with  MN  and  PQ  re- 
spectively Then  are  AC  and  BD  parallel  (?). 
Now,  AB  is  perpendicular  to  BD  (?),  and  hence 
to  AC  (?).  Thus,  AB  is  shown  to  be  perpendic- 
ular to  any  line  of  MN  passing  through  its  foot, 
and  hence  perpendicular  to  MN  (0-     Q-  e.  d. 


Fig.  233. 


STRAIGHT   LINES    AND    PLANES, 


221 


PROPOSITION    XX. 

491.  Theorem. — TJvrough  any  point  without  a  plane, 
one  plane  can  be  passed  parallel  to  the  given  plane,  and 
only  one. 

Demonstration. 

Let  MN  be  a  plane,  and  B  any  point  without  MN. 

Let  BA  be  a  perpendicular  from  B 
upon  MN. 

Through  B  draw  DE  and  FG  per- 
pendicular to  AB.  Then  is  the  plane  of 
DE  and  and  FG  parallel  to  MN  (452, 
487).    Q.  E.  D. 


Again,  as  any  plane  parallel  to  MN 
is  perpendicular  to  AB,  and  as  only  one 
plane  can  be  passed  through  B  perpen- 
dicular to  AB  (457),  only  one  plane  can 
be  passed  through  B  parallel  to  MN.     Q.  e.  d. 


Fig.  234. 


PROPOSITION     XXI. 

492.  Theorem. — Two  angles  lying  in  different  planes, 
but  having  their  sides  parallel  and  extending  in  the 
same  direction,  or  in  opposite  directions,  are  equal,  and 
their  planes  are  parallel. 

Demonstration. 


Let  A  and  A'  lie  In  the  different  planes 
MN  and  PQ,  and  have  AB  parallel  to  AB  , 
and  AC  to  AC. 

Then  A  =  A',  and  MN  and  PQ  are 
parallel. 

For,  take  AD  =  A'D',  and  AE  =  A'E', 
and  draw  AA',  DD',  EE',  ED,  and  E'D'. 
Now,  AD  being  equal  and  parallel  to 
AD', 

AA'  =  DD'  (?) 


Fig.  235. 


222 


ELEMENTARY    GEOMETRY. 


EE'; 

Again,  since  EE' 


For  like  reason, 

AA'  = 
therefore  EE'  =  DD'. 
and  DD'  are  respectively  parallel  to  AA', 
they  are  parallel  to  each  other  (?) ;  whence 
EDO'E'  is  a  parallelogram  (?),  and  ED  = 
E'D'.  Hence  the  triangles  ADE  and  A'D'E' 
are  mutually  equilateral,  and  A,  opposite 
ED,  is  equal  to  A',  opposite  E'D',  equal  to 
ED.     Q.  E.  D. 

Again,  the  plane  of  the  angle  BAC,  MN, 
is  parallel  to  PQ,  the  plane  of  B'A'C. 
For,  let  a  plane  be  passed  through  A  and  revolved  until  it  is  parallel 
to  PQ.  It  must  cut  DD'  which  is  parallel  to  AA',  and  EE'  which  also  is 
parallel  to  AA',  so  that  DD'  and  EE'  shall  equal  AA'  (?);  hence  it  must 
pass  through  D.     Hence  the  planes  of  the  angles  are  parallel,     q.  e.  d. 


493.  Corollary  l.—If  two  inter sectiiig  planes  are  cut 
by  parallel  planes,  the  angles  formed  hy  the  intersections 
are  equal. 

Thus,  AS'  and  AC  being  cut  by  the  parallel  planes  MN  and  PQ.  AD  is 
parallel  to  A'D'  (0,  and  extends  in  the  same  direction  from  vertex  A  that 
A'D'  does  from  A';  and  the  same  may  be  said  of  AC  and  A'C.  Hence, 
BAC  =  B'A'C  (?). 


494.  Corollary  2. — //  the  corresponding  extremities 
of  three  equal  parallel  lines  not  in  the  same  plane  are 
joined,  the  triangles  formed  are  equal,  and  their  planes 
parallel. 

Thus,  if  AA'  =  DD'  =  EE',  the  sides  of  the  triangle  AED  are  equal 
to  the  sides  of  A'E'D',  since  the  figures  AD',  DE',  and  EA'  are  parallelo- 
grams (?),  and  the  corollary  comes  under  the  proposition  {i). 


STRAIGHT   LINES    AND    PLANES, 


PROPOSITION    XXII. 

495.  Theorem. — //  two  lines  are  cut  by  three  parallel 
planes,  the  corresponding  intercepted  segments  are  propor- 
tional. 

Demonstration. 

Let  AB  and  CD  be  cut  by  the  three  parallel  planes  M,  N,  and  P,  AB 
piercing  the  planes  in  A,  E,  and  B,  and  CD  in  C,  F,  and  D. 

^      .    AE      CF 
^^^^^EB  =  FD- 

Join  the  points  A  and  D  by  the 
straight  line  AD,  and  eoneeive  planes 
passing  through  AD  and  DC,  and 
through  AB  and  AD. 

Let  EH  and  BD  be  the  intersec- 
tions of  the  planes  N  and  P  with  the 
plane  BAD,  and  AC  and  HF  the  in- 
tersections of  M  and  N  with  ADC- 

Now,  since  EH  is  parallel  to  BD  (?), 

AE  _  AH 

EB  ~  HD  ^'' " 


Fig.  236. 


In  like  manner,  by  reason  of  the  parallelism  of  HF  and  AC, 


CF_  AH 
FD  ~  HD' 


Hence,  by  equality  of  ratios, 


AE       CF 

Ei  =  FD-    ^"•^- 


[Note. — Planes  perpendicular  or  oblique  to  each  other  give  rise  to  one 
species  of  solid  angles;  hence  their  consideration  is  reaeired  for  the  next 
Section.] 


234  ELEMENTARY    GEOMETRY, 


EXERCISES. 

496.  1.  Designate  any  three  points  in  the  room,  as  one  cor- 
ner of  the  desk,  a  point  on  the  stove,  and  some  point  in  the 
ceiling,  and  show  how  you  can  conceive  the  plane  of  these  points. 

2.  Show  the  position  of  two  lines  which  will  not  meet,  and 
yet  are  not  parallel. 

3.  Conceive  two  lines,  one  line  in  the  ceiling  and  one  in  the 
floor,  which  shall  not  be  parallel  to  each  other. 

4.  The  ceiling  of  my  room  is  10  feet  above  the  floor.  I  have 
a  12-foot  pole,  by  the  aid  of  which  I  wish  to  determine  a  point 
in  the  floor  directly  under  a  certain  point  in  the  ceiling.  How 
can  I  do  it  ? 

Suggestion.— Consult  Proposition  XII. 

5.  Upon  what  principle  in  this  section  is  it  that  a  stool  with 
three  legs  always  stands  firm  on  a  level  floor,  when  one  with  four 
may  not  ? 

6.  By  the  use  of  two  carpenter's  squares  you  can  determine  a 
perpendicular  to  a  plane.    How  is  it  done  ? 

7.  If  you  wish  to  test  the  perpendicularity  of  a  stud  to  a  level 
floor,  on  how  many  sides  of  it  is  it  necessary  to  measure  the 
angle  which  it  makes  with  the  floor?  By  applying  the  right 
angle  of  the  carpenter's  square  on  any  two  sides  of  the  stud,  to 
test  the  angle  which  it  makes  with  the  floor,  can  you  determine 
whether  it  is  perpendicular  or  not  ? 

8.  If  a  line  is  drawn  at  an  inclination  of  23°  to  a  plane,  what 
is  the  greatest  angle  which  any  line  of  the  plane,  drawn  through 
the  point  where  the  inclined  line  pierces  the  plane,  makes  with 
the  line  ?  Can  you  conceive  a  line  of  the  plane  which  makes  an 
angle  of  50°  with  the  inclined  line  ?  Of  80°  ?  Of  15°  ?   Of  170°  ? 


SOLID    ANGLES,  225 


^^^gcfToM  \l 


OF    SOLID    ANGLES. 

497.  •  A  Solid  Angle  is  the  opening  between  two  or  more 
planes,  each  of  which  intersects  all  the  others.  The  lines  of  in- 
tersection are  called  Edges,  and  the  planes,  or  the  portion  of 
the  planes  between  the  edges  where  there  are  more  than  two, 
are  called  Faces. 

498.  Solid  Angles  are  of  Three  Species,  viz.,  Diedral, 
Triedral,  and  Polyedral,  according  as  they  have  two,  three, 
or  more  than  three  faces. 


OF     DIEDRALS. 

499.  A  Diedral  Angle,  or  simply  a  Diedral,  is  the 

opening  between  two  intersecting  planes. 

600.  A  Diedral  (Angle)  is  Measured  by  the  plane 
angle  included  by  lines  drawn  in  its  faces  from  any  point  in  the 
edge,  and  perpendicular  thereto. 

A  diedral  angle  is  called  Right,  Acute,  or  Obtuse, 
according  as  its  measure  is  right,  acute,  or  obtuse. 

Two  diedrals  are  said  to  be  Supi)leinentary,  when  their 
measures  are  supplementary. 

Of  course  the  magnitude  of  a  solid  angle  is  independent  of  the  dis- 
tances to  which  the  edges  may  chance  to  be  produced. 

Illustrations. — The  opening  ])etween  the  two  planes  CABF  and 
DABE  (Fig.  237)  is  a  Diedral  (angle),  AB  is  the  Edge,  and  CABF  and 
DABE  are  the  Faces.  Let  MO  lie  in  the  plane  AF,  perpendicular  to  the 
edge ;  and  NO  in  AE,  and  also  perpendicular  to  the  edge ;  then  the  plane 
angle  MON  is  the  measure  of  the  diedral. 


ELEMENTARY    GEOMETRY. 


Fig.  237. 


Fig.  238. 


Fig  239. 


501.  A  diedral  may  be  read  by  the  letters  on  the  edge,  when 
there  would  be  no  ambiguity,  or  otherwise  by  these  letters  and 
one  in  each  face. 

Thus,  the  diedral  in  Fig.  237  may  be  designated  as  AB,  or  as  C-AB-D. 


502.  A  diedral  may  be  considered  as  generated  by  the  revolu- 
tion of  a  plane  about  a  line  of  the  plane,  and  hence  we  may  see 
the  propriety  of  measuring  it  by  the  angle  included  by  two  lines 
in  its  faces  perpendicular  to  its  edge,  as  stated  in  the  preceding 
article. 

Illustration. — Let  AB  (Fig.  338)  be  a  line  of  the  plane  GB.  Con- 
ceive ^B  perpendicular  to  A'B.  Now,  let  the  plane  revolve  upon  AB  as 
an  axis,  whence  grB  describes  a  circle  (?)  ;  and  at  any  position  of  the  re- 
volving plane,  as  /BAF,  since  f^g  measures  the  amount  of  revolution,  it 
may  be  taken  as  the  measure  of  the  diedral  f-E^-g.  When  gB  has  made 
^  of  a  revolution,  the  plane  will  have  made  ^  of  a  revolution,  and  the 
diedral  will  be  right. 

503.  When  two  planes  intersect,  four  diedrals  are  formed,  any 
two  of  which  are  either  At\jaceiit  to  each  other,  or  Opposite. 

504.  Adjacent  Diedrals  are  on  the  same  side  of  one 
plane,  but  on  the  opposite  sides  of  the  other. 

As  D-AB-C  and  D-AB-c,  or  c-AB-D  and  c-AB-d  (Fig.  239). 

Opposite  Diedrals  are  on  opposite  sides  of  both  planes. 
As  D-AB-C  and  ^AB-c,  or  D-AB-c  and  <^AB-C  (Fig.  239). 


SOLID    ANGLES. 


227 


PROPOSITION    I. 

506.  Theorem. —  W%en  two  planes  intersect,  the  op- 
posite diedrals  are  equal,  and  the  adjacent  ones  are 
supplementary. 

Demonstration. 
Let  the  planes  DE  and  CF  intersect  in  AB. 
Then  D-AB-C  =  d-M-c, 

and  D-AB-r  =  <?-AB-C; 

and  D-AB-C +  D-AB-C*  =  180°, 

c-AB-D  +  c-AB-</  =  180°,  etc. 

Through  0,  any  point  in  AB,  let  Mm  J>e  drawn 
in  the  plane  CF,  and  Nw  in  the  plane  DE,  each 
perpenrlicular  to  AB.  Tlien  is  MON,  the  mcHsure 
of  D-AB  C  (?),  =  mOn,  the  measure  of  d-hB-c  (?), 
etc.      Q.  E.  D. 

Also,  MON  +  NOm  =  180°  (?), 

H0m-\-7n0n  =  180°,  etc. 


Fig.  240. 


q.  E.  D. 


PROPOSITION    II. 

606.  Theorem.—^/z,^  Une  in  one  face  of  a  right  die- 
dral,  perpendicular  to  its  edge,  is  perpendicular  to  the 
other  face. 

Demonstration. 

in  the  face  CB  of  the  right  diedral  C-AB-D^  let 
MO  be  perpendicular  to  the  edge  AB. 

Then  is  MO  perpendicular  to  the  face  DB. 

For,  draw  ON  in  the  face  DB,  and  perpendicu- 
lar to  AB.  Now,  since  the  diedral  is  right,  and 
MON  measures  its  angle,  MON  is  a  right  angle; 
whence  MO  is  perpendicular  to  two  lines  of  the 
plane  DB,  and  consequently  perpendicular  to  the 
plane.     Q.  e.  d. 


Fig.  241. 


*"■  By  this  is  meant  the  measure  of  the  diedral. 


228 


ELEMENTARY    GEOMETRY. 


607.  OoKOLLARY  1. — Conversely,  If  one 
plane  contains  a  line  which  is  perpen- 
dicular to  another  plane,  the  dtiedral  is 
right. 

Thus,  if  MO  is  perpendicular  to  the  plane 
DB,  C-AB-D  is  a  right  diedral.  For  MO  is  perpen- 
dicular to  every  line  of  DB  passing  through  its 
foot  (?) ;  and  hence  is  perpendicular  to  ON,  drawn 
at  right  angles  to  AB.  When  C-AB-D  is  a  right 
diedral,  for  it  is  measured  by  a  right  plane  angle.  ^'9-  ^*'- 

608.  Two  planes  are  Perpendicular  to  each  other  when 
they  intersect  so  as  to  make  the  adjacent  diedrals  equal.  In  this 
case,  all  four  of  the  diedrals  are  right. 

609.  Corollary  2.— The  plane  which  projects  a  line 
upon  a  plane  (468)  is  perpendicular  to  the  plane  of  projec- 
tion. 


PROPOSITION    III. 


610.  Theorem. — //  each  of  two  intersecting  planes  is 
perpendicular  to  a  third,  their  intersection  is  perpendicu- 
lar to  the  third  plane. 

Demonstration. 

Let  EF  3"'*  CD  bo  twu  planes  perpendicular  to  the  third  plane  MN, 
and  J^yt  AB  be  the  intersection  of  EF  and  CD. 

Then  is  AB  perpendicular  to  MN. 

For,  EF  being  perpendicular  to  MN, 
D-FG-E  is  a  right  diedral,  and  a  line  in  EF 
perpendicular  to  FG  at  B  is  perpendicular  to 
MN ;  also  a  line  in  the  plane  CD,  and  perpen- 
dicular to  DH  at  B,  is  perpendicular  to  MN  (?). 

Hence,  as  there  can  be  one  and  only  one 
perpendicular  to  MN  at  B,  and  as  this  perpen- 
dicular is  in  both  planes,  CD  and  EF,  it  is 
their  intersection,    q.  e.  d. 


Fig.  242. 


SOLID     ANGLES. 


229 


PROPOSITION    IV. 

511.  Theorem.— me  angle  Uicluded  by  perpendicu- 
lars drawn  from  any  point  within  a  diedral  to  its  faces, 
is  the  supplement  of  the  diedral. 

Demonstration. 


Fig  243. 


Let  P'  be  any  point  within  the  diedral  F-AB-C,  and  let  the  perpendic- 
ulars P'D'  and  P'E'  be  drawn  to  the  faces. 

Then  is  D'P'E'  the  sup- 
plement of  F-AB-C. 

From  P,  any  point  in 
the  plane  which  Insects  the 
diedral  F-AB-C,  draw  PD 
and  PE  perpendicular  to 
the  same  faces  respectively 
as  P'D'  and  P'E'.  Then  is 
DPE  =  D'P'E'. 

Now  pass  a  plane 
through  PE  and  PD,  and 
let  EG  and  DO  be  its  inter- 
sections with  FB  and  CB 
respectively.      Then,      by 

(507),  FB  and  CB  are  perpendicular  to  the  plane  PEOD.  Hence,  A5  is 
perpendicular  to  PEOD  (?),  and  EOD  is  the  measure  of  F-AB-C  (?).  But 
in  the  quadrilateral  PEOD,  P  is  the  supplement  of  EOD  (?),  and  hence  of 
F-AB-C. 

Hence,  D'P'E'  is  the  supplement  of  F-AB-C.     Q.  e.  d. 

512.  Corollary  1. — If  from,  a  point  in  the  edge  of  a 
diedral  perpendiculars  are  erected  to  the  faces  on  the  same 
sides  of  the  planes  respectively  as  the  perpendiculars  let 
fall  from  a  point  within,  the  included  angle  is  the  sup- 
plem^ent  of  the  angle  of  the  diedral. 

513.  Corollary  2. — The  angle  DPE  is  the  supplement 
of  the  opposite  diedral  H-AB-I,  and  equal  to  each  of  the  ad- 
jacent diedrals  C-AB-I  and  F-AB-H- 


230 


ELEMENTARY     GEOMETRY, 


PROPOSITION    V. 

514.  Theorem. — Between  any  two  lines  not  in  the 
same  plane  one  line,  and  only  one,  can  he  drawn  which 
shall  he  perpendicular  to  both,  and  this  line  is  the  shortest 
distance  between  them. 


Demonstration. 

Let  AB  and  CD  be  two  lines  not  in  the  same  plane. 

Then  one  line,  as  HG,  and  only 
one,  can  be  drawn  which  is  perpen- 
dicular to  both  AB  and  CD,  and  HG 
measures  the  shortest  distance  be- 
tween AB  and  CD. 

Through  either  line,  as  CD,  pass 
a  plane  MN  parallel  to  AB  (480). 
From  any  point  in  AB,  as  E,  let  fall 
EF  perpendicular  to  MN. 

Let  EK  be  the  plane  of  the  lines 
EF  and  EB,  and  let  FK  be  its  trace 

Fig. 

m  MN. 

Now,  as  AB  and  CD  are  not  in  the  same  plane,  EK,  and  hence  its 
trace  FK,  cuts  CD  in  some  point,  as  G. 

From  G  draw  GH  perpendicular  to  AB. 

laf    nu  i;--  :->  the  plane  EK  (?)  widch  is  perpendicular  to  MN  (?),  and 
being  perpendicular  to  AB  is  perpemUcuIar  to  FK  0  ,  and  hence  to  the 

GH,  which  is  perpendicular  to  AB,  is  perpendicular  to 


244. 


■ia.  an  IS  the  only  line  which  is  perpendicular  U)    oth  AB  and  CD. 

For  ftny  line  which  is  prrpendicular  to  A5  aua  C^  is  perpendicular 
to  FK  (?),  and  hence  to  MN  (?). 

Now  every  perpendicular  from  AB  to  the  plane  MN  meets  this  plane 
in  FK  (?). 

But  FK  and  CD  have  only  one  point  common,  viz.,  G.  Hence,  GH  is 
the  only  perpendicular  from  AB  to  CD. 

3d.  GH  is  the  shortest  distance  between  AB  and  CD.  For  a  line  from 
any  point  in  AB  to  any  other  point  in  CD,  as  LS,  would  be  oblique  to 
MN  (?),  and  hence  longer  than  the  perpendicular  LR,  =  HG. 


BOLID    ANGLES, 


231 


PROPOSITION    VI. 

515.   Theorem. — //  one  of  two  parallel  planes  is  per- 
pendicular to  a  third  plane,  the  other  is  also. 

Demonstration. 


Let  PO  and  QE  be  two  paral 
lei  planes;  and  let  PD  be  perpen- 
dicular to  the  third  plane  MN. 

Then  is  QE  perpendicular  to 
MN. 

Through  PD  and  QE  pass  the 
plane  RS  perpendicular  to  MN, 
and  let  FK  be  its  trace  in  QE,  and 
HI  in  PD. 

Then  is  FK  perpendicular  to 
MN  (?). 

And,  as  HI  is  parallel  to  FK 
(?),  it  is  perpendicular  to  MN 
(460). 

Hence,  QE  is  perpendicular 
to  MN  (507).    Q.  K.  D. 


Fig.  245. 


OF    TRIEDRALS. 

516.   As  diedrals  result  from  the  intersection  of  two  planes, 
so  triedrals  result  from  the  intersection  of  three  planes. 


Fifl.  24d. 


232  ♦  ELEMENTARY    GEOMETRY. 

517.  Three  planes  may  intersect  in  three  principal  ways: 

1st.  Their  intersections  may  all  coincide,  as  in  (a). 

2d.  They  may  have  three  parallel  intersections,  as  in  (h), 

3d.  They  may  have  three  non-parallel  intersections,  as  in  (c). 
In  this  case  the  three  intersections  meet  in  a  common  point, 
as  at  S. 

In  the  first  case  the  three  planes  have  an  infinite  number  of 
common  points.  In  the  second  case  they  have  no  common  point. 
In  the  third  case  they  have  but  one  common  point. 

The  third  case  gives  rise  to  Triedrals. 

618.  A  Triedral  is  the  opening  between  three  planes  which 
meet  in  a  common  point. 

519.  When  three  planes  meet  so  as  to  form  one  triedral,  they 
form  also  eight,  as  planes  are  to  be  considered  indefinitely  ex- 
tended, unless  otherwise  stated. 

520.  The  planes  enclosing  a  particular  triedral  are  called  its 
Faces,  and  their  intersections  its  Edges.  The  common  point 
is  called  the  Vertex. 

521.  A  triedral  may  be  designated  by 
naming  the  letter  at  the  vertex  and  then 
three  other  letters,  one  in  each  edge. 

Thus,  in  the  figure,  the  opening  between  the 
three  planes  ASC,  CSB,  and  BSA  is  the  triedral 
S-ABC.     ThQ  faces  are  ASC,  CSB,  and  BSA. 

Fig.  247. 

522.  The  plane  angles  enclosing  a  solid  angle  are  called 
Facial  Angles. 

523.  In  every  particular  triedral  there  are  six  parts,  Three 
Facial  Angles  and  Three  Diedrals. 


SOLID    ANGLES, 


233 


524.  Our  study  of  triedrals  will  be  confined  to  the  relations 
of  the  facial  angles  and  the  diedrals,  and  the  comparison  of  dif- 
ferent triedrals. 


525.  Triedrals  are  Rectangular,  Bi-rectang^ular,  or 

Tri-rectangular,  according  as  they  have  one,  two,  or  three 
right  diedral  angles. 

Illubtbation. — The  comer  of  a  cube  is  a 
Tinrectcmgular  triedral,  as  S-ADC.  Conceive  the 
upper  portion  of  the  cube  removed  by  the  plane 
ASEF ;  then  the  angle  at  S,  i.  e.,  S-AEC,  is  a  Bi- 
rectangular  triedral,  A-SC-E  and  A-SE-C  being 
right  diedrals.  Pig.  248. 

626.  An  Isosceles  Triedral  is  one  that  has  two  of  ito 
facial  angles  equal.  An  Equilateral  Triedral  is  one  that 
has  all  three  of  its  facial  angles  equal. 

527.  Opposite  Triedrals  are  such  as  lie  on  opposite  sides 
of  each  of  the  intersecting  planes,  as  S-ABC  and  %-ahc. 

Opposite  triedrals  have  mutually  equal  facial  and  equal 
diedral  angles,  but  these  being  differently  disposed,  such 
triedrals  are  not  in  general  capable  of  superposition. 

Illustration. — Let  the  edges  of  the  triedral  S-ABC  be 
produced  beyond  the  vertex,  forming  the  opposite  triedral 
S-aZw.  Now,  the  faces  are  equal  plane  angles,  but  disposed 
in  a  different  order.  Thus,  ASB  =  aS&,  ASC  =  aSc,  and 
BSC  =  6Sc,  and  the  diedrals  are  also  equal ;  but  the 
triedrals  cannot  be  superimposed,  or  made  to  coincide.  To 
show  this  fact,  conceive  the  upper  triedral  detached,  and 
the  face  aSc  placed  in  its  equal  face  ASC,  Sa  in  SA,  and  Sc  ^'9-  ^*^- 
in  SC.  Now  the  edge  S6,  instead  of  falling  in  SB,  in  front  of  ASC,  will 
fall  behind  the  plane  ASC. 

Or,  otherwise,  if  %-ahc  be  revolved  on  S  by  bringing  it  forward  and 
turning  it  down  on  S-ABC,  since  the  diedrals  A-SB-C  and  o-Sh-a  are 
equal,  they  will  coincide ;  but,  as  facial  angle  aSh  is  not  necessarily  equal 
to  CSB,  S/i  will  not  necessarily  fall  in  SC.  For  a  like  reason,  %c  will  not 
necessarily  fall  in  SA. 


234 


ELEMENTARY    GEOMETRY, 


528.  Symmetrical  Triedrals  are  triedrals  in  which  each 
part  in  one  has  an  equal  part  in  the  other ;  but  the  equal  parts 
not  being  similarly  disposed,  the  triedrals  may  not  be  capable  of 
superposition. 

Symmetrical  solids  are  of  frequent  occurrence :  the  two  hands  form 
an  illustration ;  for,  though  the  parts  may  be  exactly  alike,  the  hands 
cannot  be  placed  so  that  their  like  parts  "will  be  similarly  situated ;  in 
short,  the  left  glove  will  not  fit  the  right  hand. 

529.  Atljaceiit  Triedrals  are  such  as  lie  on  different 
sides  of  one  of  the  intersecting  planes,  and  on  the  mme  side  of 
two  of  them. 


Thus,  S-ADE  is  adjacent  to 
8.DRE. 

In  adjacent  triedrals,  two  of 
the  facial  angles  of  one  are  the 
supplements  of  two  of  the  other, 
each  to  each,  and  one  is  equal  in 
each 

Thus,  in  the  adjacent  triedrals 
S-DRE  and  S-ADE,  ASE  and  ASD 
are  supplements  respectively  of  ESR 
and  DSR,  while  DSE  is  common  to 
both. 


Fig.  250. 


530.  Of  the  eight  triedrals  formed  by  the  intersection  of 
three  planes,  each  has  its  Oi)i)Osite  or  Symmetrical 
triedral,  and  each  has  three  Adjacent  triedrals. 

631.  Two  triedrals  are  Supplementary  when  the  facial 
angles  of  the  one  are  the  supplements  of  tiie  measures  of  the 
corresponding  diedrals  of  the  other. 


532.  Equality,  as  has  been  before  defined,  means,  in  Geom- 
etry, equality  in  all  respects  ;  and  two  figures  that  are  said  to  be 
equal  are  capable  of  being  so  applied  the  one  to  the  other  that 
they  will  coincide  throughout.     This  absolute  equality  is  hence 


SOLID    ANGLES, 


235 


often  called  Equality  by  Superposition^  in  distinction 
from  Equality  by  Symmetry. 

533.  Two  figures  are  said  to  be  Equal  by  Symmetry,  or 
Symmetrically  Equals  or  simply  Symmetrical,  when 
each  part  in  one  has  an  equal  part  in  the  other ;  but  these  equal 
parts  being  differently  arranged  in  the  two  figures,  the  one  may 
not  be  capable  of  being  superimposed  upon  the  other.  (See  527.) 


PROPOSITION    VII. 

534.  Theorem. — Opposite  triedrals  are  equal  and  may 
be  syninietrlcal. 

Demonstration. 

Let  S-ABC  and  S-ahc  be  two  opposite  triedrals. 

Then  are  the  triedrals  equal  or  syiiiiiietricul. 

For  the   facial  angle  ASC  =  the  facial  angle  aSc  (?); 
also,  BSC  =  bSc,  and  ASB  =  aSh. 

Again,  the  diedra'  A-SB-C  =  a-Sft-c,  since  they  are  op- 
posite diedrals. 

For  like  reason,  B  SA-C  =  b-Sa-c,  and  A-SC-B  =  a-Sc-b. 

Hence  all  the  parts  in  one  triedral  have  equal  parts  in 
the  other. 

But,  in  general,  these  triedrals  cannot  be  superimposed. 
(See  illustration,  527.)  '** 

If,  however,  ASB  =  CSB,  then  aSb  =  cSb,  and  the  triedrals  can  be 
superimposed. 

Thus,  conceive  the  triedral  S-ahc  revolved  on  S,  being  brought  over 
towards  the  observer  until  Sb  falls  in  SB. 

Tiien,  since  CSB  =  ASB  =  aSb,  aSb  may  be  made  to  coincide  with 
BSC,  and  as  the  diedrals  A-SB-C  and  ct-Sb-c  are  equal,  cSb  will  fall  in 
ASB,  and  the  triedrals  will  coincide,  and  will  be  equal. 

Hence,  opposite  diedrals  are  equal  and  may  be  symmetrical,     q.  e.  d. 


535. 
equal. 


Corollary  1.  —  Opposite    isosceles   triedrals   are 


236 


ELEMENTABY    GEOMETRY. 


PROPOSITION    VIII. 

536.  Theorem. — Two  syrmnetrical  triedrals  may  al- 
ways be  conceived  to  be  placed  as  opposite  triedrals. 


Demonstration. 

Let  S-ABC  and  S'-A'B'C  be  two  symmetrical  triedrals,  B  and  B' 
being  in  front  of  the  planes  ASC  and  A'S'C,  ASB  =  A'S'B',  ASC  = 
A'S'C,  BSC  =  B'S'C',  A-SB-C  =  A'-S'B-C',  A-SC-B  =- A'-SX'-B',  and 
B-SA-C  =  B'-S'A'-C. 

Then  may  S-ABC  and  S'-A'B'C 

be  placed  as  opposite  triedrals. 

Produce  the  edges  of  either 
triedral,  as  S'-A'B'C,  beyond  the 
vertex,  forming  the  opposite  tri- 
edral S'-aftc. 

Then  can  S-ABC  be  super- 
imposed upon  ^'-dbc,  and  the  latter 
fulfills  the  requirements  of  the 
proposition. 

The  application  is  made  as 
follows : 

Since  B'  is  in  front  of  the  plane  Fig.  252. 

A'S'C,  I  is  behind  the  plane  aS'c. 

Now  conceive  S-ABC  inverted  and  reversed  so  that  B  shall  fall 
behind  the  plane  ASC. 

Then  apply  ASC  to  its  equal  aS'c,  SA  falling  in  S'a,  and  SC 
in  S'c. 

By  reason  of  the  equality  of  A-SC-B  and  a-S'c-&  (=  A'-S'C'-B'),  the 
plane  BSC  will  fall  in  5S'c,  and  for  a  like  reason  ASB  will  fall  in 
aS'5 ;  and  since  the  planes  coincide,  their  intersections  SB  and  S'ft  must 
coincide. 

Hence,  S-ABC  =  S'-a5c,  the  opposite  to  S'-A'B'C     Q.  E.  d. 


SOLID    ANGLES, 


237 


PROPOSITION    IX. 

537.  Theorem. — Two  triedrals  which  have  two  facial 
angles  and  the  included  diedral  equal,  each  to  each,  are 
either  equal  or  symmetrical, 

Demonsteation. 

Let  I,  2,  3,  be  triedrals  having  the  facial  angle  ASC  =  A'S'C  =  aS"c, 
CSB  =  C'S'B'  =  cS "6,  and  A-SC-B  =  A -S'C'-B'  =  a-^'c-h. 


Fig.  253. 


Then  are  the  triedrals  either  equal  or  symmetrical. 


Ist.  When  the  equal  facial  angles  are  on  the  same  sides  of  the  respec- 
tive equal  diedrals,  as  in  Figs.  2  and  3,  the  triedrals  may  be  appliecf  the 
one  to  the  other. 

Thus,  let  the  facial  angle  A'S'C  be  placed  in  its  equal  oS"c,  A'S'  in 
aS,  and  S'C  in  S"c;  whence,  by  reason  of  the  equality  of  the  diedrals 
A'-S'C'-B'  and  a-S"c-6,  and  since  the  facial  angles  B'S'C  and  hS"c  lie  on 
the  same  sides  respectively  of  their  diedrals  A'-S'C  -B'  and  a-S"c-h,  the 
plane  of  B'S'C  falls  in  the  plane  of  6S"c,  and  since  angle  B'S'C  =  angle 
6S"c,  B'S'  falls  in  6S",  and  A'S'B'  coincides  with  aS"&. 

Hence  the  triedrals  coincide  and  are  equal,    q.  e.  d. 

2d.  But  if  the  equal  facial  angles  lie  on  different  sides  of  the  equal 
diedrals,  as  in  Figs.  1  and  3,  let  the  opposite  of  S-ABC  be  drawn  (627), 
and  call  it  S-a'Vcf.     Then  may  1  be  applied  to  S-a'h'd. 

[Let  the  student  draw  the  figure  and  make  the  application.] 


338 


ELEMENTARY    GEOMETRY. 


PROPOSITION    X. 

538.  Theorem. — Two  triedrals  which  have  two  die- 
drals  and  the  included  facial  angles  equal  each  to  each, 
are  either  equal  or  symmetrical, 

Demoi^stbation. 
[Same  as  preceding.     Let  the  student  draw  figures  like  those  for  the 
preceding,  and  go  through  with  the  details  of  the  application.] 

539.  Corollary. — In  equal  or  in  symmetrical  triedrals, 
the  equal  facial  angles  are  opposite  the  equal  diedrals. 


PROPOSITION    XI. 

540.  Theorem. — The  sum  of  any  two  facial  angles  of 
a  triedral  is  greater  than  the  third- 

Demonstration. 
This  proposition  needs  demonstration  only  in  case  of  the  sum  of  the 
two  smaller  facial  angles  as  compared  with  the  greatest  {(). 
Let  ASB  and  BSC  each  be  less  than  ASC ;  then  is 

ASB  +  BSC  >  ASC. 

For,  in  the  face  ASC,  make  the  angle  AS  J'  =  ASB, 
and  S5'  =  S&,  and  pass  a  plane  through  5  and  6',  cut- 
ting SA  and  SC  in  «  and  c. 

The  two  triangles  aSh  and  a%V  are  equal  (?),  whence 

Now,  ab  ■\-'bc>  a^  (?),  Fig,  254. 

and  subtracting  ah  from  the  first  member,  and  its  equal  cikl  from  the  sec- 
ond, we  have  he  >  ft'c. 

Whence  the  two  triangles  5Sc  and  V%c  have  two  sides  in  the  one 
equal  to  two  sides  in  the  other,  each  to  each,  but  the  third  side  he  >  than 
the  third  side  ft'c,  and  consequently  angle  BSC  >  h  SC.     Adding  ASB  to 
the  former,  and  its  equal  AS&'  to  the  latter,  we  have 
ASB  +  BSC  >  ASC.     Q.  E.  D. 

541.  Corollary. — The  difference  hetween  any  two  facial 
angles  of  a  triedral  is  less  than  the  third  facial  angle  (?). 


SOLID    ANGLES.  239 


PROPOSITION    XII. 

542.  Theorem. — Two  triedrals  which  have  two  facial 
angles  of  the  one  equal  to  two  facial  angles  of  the  other, 
each  to  each,  and  the  included  diedrals  unequal,  have  the 
third  facial  angles  unequal,  and  the  greater  facial  angle 
belongs  to  the  trledral  having  the  greater  included  diedi'al. 

Demonstr'ation. 

Let  ASC  =  asc,  and  ASB  =  asb,  while  the 
diedral  C-SA  B  >  c-sa-b. 

Then  CSB  >  eab. 

For,  divide  the  diedral  C-SA-B  by  a  plane 
ASO,  making  the  diedral  C-SA-0  =  o-sa-b; 
end  taking  ASO  =  aid,  bisect  the  diedral 
0-SA-B  with  the  plane  ISA.  Conceive  the 
planes  OSI  and  OSC.  F«g-  255. 

Now,  the  triedrals  S-AOC  and  s-ahc  are  equal  or  symmetrical,  having 
two  facial  angles  and  the  included  diedral  equul  eael^^to  each  (637). 

For  a  like  reason,  S-AIO  and  S-AIB  are  symmetrical,  and  the  facial 
angle  OSI  =  ISB. 

Again,  in  the  triedral  S-IOC> 

OSI  +  ISC  >  OSC  (640), 
and  substituting  ISB  for  OSI,  we  have 

ISB  -I-  ISC  (or  CSB)  >  OSC,  or  its  equal  csb.    q.  e.  d. 

643.  Corollary.— Conversely,  //  tJte  two  facial  angles 
are  equal,  each  to  each,  in  two.  triedrals,  and  the  third 
facial  angles  unequal,  the  diedral  opposite  the  greater 
facial  angle  is  the  greater. 

That  is,  if  ASB  =  asb,    and     ASC  =  anCy 

while  BSC  >  bsc, 

the  diedral  B-AS-C  >  b-as-r. 

For,  if  B-AS-C  ^  b-as-c,        BSC  =  bse  (637,  639) ; 

and  if  B-AS  C  <  b-aa-c,        BSC  <  bse,  by  the  proposition. 

Therefore,  as  B-AS-C  cannot  be  equal  to  nor  less  than  b-as-e,  it  must 
be  greater,    q.  e.  d. 


240 


ELEMENTARY    OEOMETRT, 


PROPOSITION     XIII. 

M  544.  Theorem. — Two  triedrals  which  have  the  three 
facial  angles  of  the  one  equal  to  the  three  facial  angles  of 
the  other,  each  to  each,  are  either  equal  or  symmetrical, 

Demokstration. 

Let  A,  B,  and  C  represent  the  facial  angles  of  one,  and  a,  5,  and  c  the 
corresponding  facial  angles  of  the  other.  If  A  =  a,  B  =  &,  and  C  :^  c, 
the  triedrals  are  equal  or  symmetrical. 

For  A  being  equal  to  a,  and  B  to  &,  if,  of  their  included  diedrals,  SM 
were  greater  than  «m,  C  would  be  greater  than  c  (?) ;  and  if  diedral  SM 
were  less  than  diedral  sm^  C  would  be  less  than  c  (?).  Hence,  as  diedral 
SM  can  neither  be  greater  nor  less  than  diedral  sm,  it  must  be  equal  to  it. 

Therefore  the  triedrals  have  two  facial  angles  and  the  included  diedral 
equal,  each  to  each,  and  are  consequently  equal  or  symmetrical.     Q.  e.  d. 


PROPOSITION    XIV. 

545.  Theorem,— ///roT^Z/  any  point  within  a  triedral 
perpendiculars  are  drawn  to  the  faces,  they  will  be  the  edges 
of  a  supplementary  triedral. 

Demonstratioit. 

From  S'  within  the  triedral  S-ABC,  let  S'A'  be  drawn  perpendicular 
to  ASB,  SB'  to  ASC,  and  S'C  to  BSC. 

Then  is  S'-A'B'C  supple- 
mentary to  S-ABC. 

For  the  facial  angle  A'S'B' 
is  the  supplement  of  the  di- 
edral B-AS-C  (511);  and  for 
like  reason  B'S'C  is  the  sup- 
plement of  A-SC-B,  andA'S'C 
of  A-SB-C. 

Again,  since  S'A'  is  per- 
pendicular to  the  face  ASB, 
and  S'B'  is  perpendicular  to 

ASC,  the  plane  of  S'A'   and  ^.     „^^ 

rig.  Z5o. 


SOLID    ANGLES.  241 

S'B'  is  perpendicular  to  ASB  and  ASC,  and  therefore  to  SA.  Hence  SA 
is  perpendicular  to  the  face  A'S'B'. 

For  a  similar  reason,  SC  is  perpendicular  to  B'S'C  Hence  ASC  is 
the  supplement  of  A'-S'B'-C 

In  like  manner,  it  may  be  shown  that  BSC  is  the  supplement  of 
A'-S'C'-B',  and  ASB  of  B'-S'A'-C.     Q.  e.  d. 

546.  Scholium  1. — If  perpendiculars  were  drawn  from  the  point  S, 
or  any  other  point,  parallel  to  those  from  S',  and  in  the  same  directions 
respectively  from  S  that  S'A',  etc.,  are  from  S',  they  would  also  l)e  per- 
pendicular to  the  faces  of  the  diedral,  and  would  form  a  supplementary 
triedral. 

547.  Scholium  2. — The  triedral  S'-A'B'C  is  also  supplementary  to 
the  triedral  opposite  to  S-ABC. 

548.  Scholium  3. — The  triedral  S'-A'B'C  will  not  be  supplementary 
to  the  triedral  adjacent  to  S-ABC,  but  one  facial  angle  will  be  supple- 
mentary to  the  corresponding  diedral  in  the  other,  and  the  other  facial 
angles  will  be  equal  to  their  corresponding  diedrals. 

549.  Scholium  4. — One  triedral  adjacent  to  S'-A'B'C  will  be  sup- 
plementary to  one  of  those  adjacent  to  S-ABC. 


PROPOSITION    XV. 

550.  Theorem. — In  an  isosceles  triedral  the  diedrals 
opposite  the  equal  facial  angles  are  equal ;  and, 

Conversely,  If  two  diedrals  of  a  triedral  are  equal,  the 
triedral  is  isosceles. 

Demon^stration. 

In  the  triedral  S-ABC,  let  ASC  =  CSB. 

Then  is  C-SA-B  =  C-SB-A. 

For,  pass  the  plane  CSD  through  the  edge 
SC,  bisecting  the  diedral  A-SC-B.  Then  the 
two  triedrals  S-ACD  and  S-CBD  have  two  facial 
angles  of  one  equal  to  two  facial  angles  of  the 
other,  each  to  each ;  that  is,  ASC  =  CSB,  by 
hypothesis,   and  CSD   common;    and   the  in-  P,     257 


242  ELEMENTARY    GEOMETRY. 

eluded  diedral8  equal  by  construction.     Hence  the   triedrals  are  sym- 
metrical, and 

C-SA-B  =  C-SB-A  (537,  639).    Q.  k.  d. 

Convereely,  if  C-SA-B  =  C-SB-A, 

ASC  =  CSB. 

For  the  supplementary  triedral  is  isosceles ;  whence  the  diedrals  op- 
posite those  equal  facial  angles  are  equal.  But  ASC  and  CSB  are  the 
supplements  of  these  equal  diedrals,  and  hence  equal,    q.  e.  d. 

651.  Corollary  1. — The  plane  which  bisects  the  angle 
included  by  the  equal  facial  afigles  of  an  isosceles  triedral 
is  perpendicular  to  the  opposite  face,  and  bisects  the  oppo- 
site facial  angle. 

662.  Corollary  2.—//  the^  three  facial  angles  of  a  tri- 
edral are  equal,  each  to  each,  the  diedrals  are  also  equal, 
each  to  each,  and  conversely. 


PROPOSITION    XVI. 

663.  Theorem. — Two  triedrals  which  have  the  three 
diedrals  of  the  one  equal  to  the  three  diedrals  of  the  other, 
each  to  each,  are  equal  or  symmetrical. 

Demonstration. 

In  the  two  supplementary  triedrals,  the  facial  angles  of  the  one  are 
equal  to  the  facial  angles  of  the  other,  each  to  each,  since  they  are  sup- 
plements of  equal  diedrals  (546).  Hence,  the  supplementary  triedrals 
are  equal  or  symmetrical  (644). 

Now,  the  facial  angles  of  the  first  triedrals  are  supplements  of  the 
diedrals  of  the  supplementary ;  whence  the  corresponding  facial  angles, 
being  the  supplements  of  equal  diedrals,  are  equal.  Therefore,  the  pro- 
posed triedrals  have  their  facial  angles  equal,  each  to  each,  and  are  con- 
sequently equal  j  or  symmetrical.     Q.  E.  D. 

664.  Corollary.  —  All  tri-rectangular  triedrals  are 
equal. 


SOLID    ANGLES,  343 

^<  V 

PROPOSITION    XVII. 

565.  Theorem. — The  sum  of  the  facial  angles  of  a 
-^    triedral  may  he  anything  between  zero  and  four  right 
angles. 

Demonstration. 

Let  ASB,  BSC,  and  ASC  be  the  facial  angles 
enclosing  a  triedral. 

Then,  as  each  must  have  some  value,  the  sum 
is  greater  than  zero,  and  we  have  only  to  show 
that  ASB  +  ASC  +  BSC  is  less  than  4  right  angles. 

Produce  either  edge,  as  AS,  to  D.  Now,  in  the 
triedral  S-BCD,  BSC  is  less  than  BSD  +  CSD  (?).  '"'s-  '^^^^ 

To  each  member  of  this  inequality  add  ASB  +  ASC,  and  we  have 

ASB  +  ASC  +  BSC  less  than  ASB  +  ASC  +  BSD  +  CSD  (?). 

But  ASB  +  BSD  =  2  right  angles  (?), 

and  ASC  +  CSD  =  2  right  angles; 

whence  ASB  4-  ASC  +  BSD  +  CSD  =  4  right  angles, 

and  consequently    ASB  +  ASC  +  BSC  is  less  than  4  right  angles,     q.  k.  d. 


PROPOSITION    XVIII. 

656.  Tlxeorem.—The  sum  of  the  diedrals  of  a  triedral 
may  he  anything  hetween  two  and  six  right  angles. 

Demonstration. 

Each  diedral  being  the  supplement  of  a  facial  angle  of  the  supple- 
mentary triedral  (531),  the  sum  of  the  three  diedrals  is  3  times  2  right 
angles,  or  6  right  angles,  minus  the  sum  of  the  facial  angles  of  the  sup- 
plementary triedral. 

But  this  latter  sum  may  be  anything  between  0  and  4  right  angles  (?). 
Hence  the  sum  of  the  diedrals  may  be  anything  between  2  and  6  right 
angles,    q.  e.  d. 


\ 


244 


ELEMENTARY    GEOMETRY. 


OF    POLYEDRALS. 

557.  A  Convex  Polyedral  is  a  polyedral  none  of  the 
faces  of  which,  when  produced,  enter  tlie  solid  angle.  A  sec- 
tion of  such  a  polyedral  made  by  a  plane  cutting  all  its  edges  is 
a  convex  polygon.     (See  Fig.  259. ) 


PROPOSITION    XIX. 

558.    Theorem. — Tke  sum  of  the  facial  angles  of  any 
convex  polyedral  is  less  than  four  right  angles. 


Demonstration. 
Let  S  be  the  vertex  of  any  convex  polyedral. 

Then  is  the  sum  of  the  angles  ASB,  BSC,  CSD, 
DSE,  and  ESA  less  than  4  right  angles. 

Let  the  edges  of  this  polyedral  be  cut  by  any 
plane,  as  ABODE,  which  section  will  be  a  convex 
polygon,  since  the  polyedral  is  convex. 

From  any  point  within  this  polygon,  as  0, 
draw  lines  to  its  vertices,  as  OA,  OB,  00,  etc. 
There  will  thus  be  formed  two  sets  of  triangles, 
one  with  their  vertices  at  S,  and  the  other  with 
their  vertices  at  0;  and  there  will  be  an  equal 
number  in  each  set,  for  the  sides  of  the  polygon 
form  the  bases  of  both  sets. 

Now,  the  sum  of  the  angles  of  each  of  these  two  sets  of  triangles  is 
the  same.  But  the  sum  of  the  angles  at  the  bases  of  the  trian«;les  having 
their  vertices  at  S  is  greater  than  the  sum  of  the  angles  at  the  bases  of 
the  triangles  having  their  vertices  at  0,  since  SBA  +  SBO  is  greater 
than  ABO,  SOB  +  SOD  is  greater  than  BOD,  etc.  (540). 

Therefore  the  sum  of  the  angles  at  S  is  less  than  the  sum  of  the  angles 
at  0>  ».  6-,  less  than  4  right  angles,     q.  e.  d. 


Fig.  259. 


i 


EXEKCISES,  246 


EXERCISES 


659.  1.  I  have  an  iron  block  whose  corners  are  all  square 
(edges  right  diedrals,  and  the  vertices  tri-rectangular,  or  right, 
triedrals).  If  I  bend  a  wire  square 
around  one  of  its  edges,  as  c  S't?,  at 
what  angle  do  I  bend  the  wire  ?  If 
I  bend  a  wire  obliquely  around  the 
edge,  as  aSh,  at  what  angle  can  I 
bend  it  ?  If  I  bend  it  obliquely,  as 
e  S"/,  at  what  angle  can  I  bend  it  ? 

2.  Fig.  260    represents   the  ap- 
pearance of  a  rectangular  parallelo-  Fi7"26o7 
piped,  as  seen  from  a  certain  position. 

Now,  all  the  angles  of  such  a  solid  are  right  angles :  why  is  it 
that  they  nearly  all  appear  oblique  ?  Can  you  see  a  right  paral- 
lelopiped  from  such  a  position  that  all  the  angles  seen  shall 
appear  as  right  angles  ? 

3.  The  diedral  angles  of  crystals  are  measured  with  great 
care,  in  order  to  determine  the  substances  of  which  the  crystals 
consist.  How  must  the  measure  be  taken?  If  we  measure 
obliquely  around  the  edge,  shall  we  get  the  true  value  of  the 
angle  ? 

4.  Prove  that  if  three  planes  intersect  so  as  to  make  two 
traces  parallel,  the  third  is  parallel  to  each  of  these. 

5.  From  a  piece  of  pasteboard  cut  two  figures  of  the  same 
size,  like  ABCDS  and  ahcds  (Fig.  261).  Then  drawing  SB  and 
SC  so  as  to  make  1  the  largest  angle  and  3  the  smallest,  cut  the 
pasteboard  almost  through  in  these  lines,  so  that  it  will  readily 
bend  in  them.  Now  fold  the  edges  AS  and  DS  together,  and  a 
triedral  will  be  formed.  From  the  piece  ahcds  form  a  triedral  in 
like  manner,  only  let  the  lines  sc  and  sh  be  drawn  so  as  to  make 


24:6  ELEMENTARY    GEOMETRY. 

the  angles  1,  2,  and  3  of  the  same  size  as  be- 
fore, while  they  occur  in  the  order  given  in 
(ibcds.  Now,  see  if  you  can  slip  one  triedral 
into  the  other,  so  that  they  will  fit.  What 
is  the  difficulty? 

6.  In  the  last  case,  if  1  equals  |  of  a  right 
angle,  2  =  ^  of  a  right  angle,  and  3  =  |  of 
a  right  angle,  can  you  form  the  triedral? 
Why  ?  If  you  keep  increasing  the  size  of  1, 
2,  and  3,  until  the  sum  becomes  equal  to  4 
right  angles,  will  it  always  be  possible  to  Fig.  26i. 
form  a  triedral  ?     How  is  it  when  the  sum  equals  4  right  angles? 

7.  What  is  the  locus  of  a  point  in  space  equidistant  from 
three  given  points  ? 

To  demonstrate  that  such  a  locus  is  a  straight  line,  pass  a  plane 
through  the  three  points,  and  also  a  circumference.  Now,  1st,  a  perpen- 
dicular to  this  circle  at  its  centre  has  every  point  equidistant  from  the 
three  points;  and,  2d,  any  point  out  of  the  perpendicular  is  unequally 
distant  from  the  points.     Hence  this  perpendicular  is  the  locus  sought. 

Notice  that  in  demonstrating  such  a  proposition  the  two  points  should 
both  be  proved. 

8.  The  locus  of  a  point  equidistant  from  two  planes  is  the 
plane  which  bisects  the  diedral  included  between  them.  [Give 
proof.] 

9.  What  is  the  locus  of  a  point  in  space  equidistant  from  the 
faces  of  a  triedral  ?     [Give  proof.] 

10.  If  each  of  the  projections  of  a  line  upon  three  intersecting 
planes  is  a  straight  line,  the  line  is  a  straight  line. 

11.  To  find  the  point  in  a  plane  such  that  the  sum  of  its  dis- 
tances from  two  given  points  without  the  plane,  and  on  the 
same  side  of  it,  shall  be  a  minimum. 

Solution. — Let  the  two  points  be  P  and  P'.  Let  fall  a  perpendicular 
from  either  point,  as  P,  upon  the  plane,  and  call  it  PD.  Produce  PD  on 
the  opposite  side  of  the  plane  to  P",  making  P"D  =  PD.  Join  P"  and  P'. 
The  point  where  P"P'  pierces  the  plane  is  the  point  sought.  [Give  proof.] 


PRISMS, 


247 


0irCtlOM  m. 


OF    PRISMS    AND    CYLINDERS. 

660.  A  Prism  is  a  solid,  two  of  whose  faces  are  equal,  par- 
allel polygons,  while  the  other  faces  are  parallelograms.  The 
equal  parallel  polygons  are  the  Bases,  and  the  parallelograms 
make  up  the  Lateral  or  Convex  Surface.  Prisms  are  triangular, 
quadrangular,  pentagonal,  etc.,  according  to  the  number  of  sides 
of  the  polygon  forming  a  base. 

561.  A  Right  Prism  is  a  prism  whose  lateral  edges  are 
perpendicular  to  its  bases.  An  Oblique  Prism  is  a  prism 
whose  lateral  edges  are  oblique  to  its  bases. 


562.  A  Regular  Prism  is  a  right  prism  whose  bases  are 
regular  polygons  ;  whence  its  faces  are  equal  rectangles. 

563.  The  Altitude  of  a  prism  is  the  perpendicular  distance 
between  its  bases  :  the  altitude  of  a  right  prism  is  equal  to  any 
one  of  its  lateral  edges. 

564.  A  Truncated  Prism  is  a  portion  of  a  prism  cut  off 
by  a  plane  cutting  the 
lateral  edges,  but  not 
parallel  to  its  base.  A 
section  of  a  prism  made 
by  a  plane  perpendicu- 
lar to  its  lateral  edges  is 
called  a  Right  Section. 

Illustrations. — In  the 
figure,  (a)  and  (&)  are  both 
prisms :  (a)  is  oblique  and 
(b)  right.  PO  represents 
the   altitude   of  (a) ;    and  '"'S-  262. 

any  edge  of  (J),  as  JB,  is  its  altitude.     ABCDEF  and  abcd^  are  lower  and 


248 


ELEMENT  A  It  Y    GEOMETR  Y. 


upper  bases,  respectively.    Either  portion  of  (b)  cut  off  by  an  oblique 
plane,  as  a'l'c'd'e',  is  a  truncated  prism. 

565.  A  Parallelepiped  is  a  prism  whose  bases  are  paral- 
lelograms ;  its  faces,  inclusive  of  the  bases,  are  consequently  all 
parallelograms.  If  its  faces  are  all  rectangular,  it  is  a  rectangu- 
lar parallelopiped. 

566.  A  Cube  is  a  rectangular  parallelopiped  whose  faces  are 
all  equal  squares. 

567.  The  Volume  or  Contents  of  a  solid  is  the  number 
of  times  it  contains  some  other  solid  taken  as  the  unit  of  meas- 
ure ;  or  it  is  the  ratio  of  one  solid  to  another  taken  as  the  stand- 
ard of  measure. 

In  applied  geometry  the  unit  of  volume  is  usually  a  cube  de- 
scribed on  some  linear  unit,  as  an  inch,  a  foot,  a  yard,  etc.  To 
this  the  perch  and  the  cord  are  exceptions. 


PROPOSITION    I. 

568.  Theorem. — Parallel  plane  sections  of  any  prism 
are  equal  polygons. 

Demoksteation^. 
Let  ABODE  and  ahcde  be  parallel  sections  of  the  prism  MN* 

Then  are  they  equal  polygons. 

For,  the  intersections  with  the  lateral  faces,  as 
ab  and  AB,  etc.,  are  parallel,  since  they  are  inter- 
sections of  parallel  planes  by  a  third  plane  (488)- 

Moreover,  these  intersections  are  equal,  that 
is,  db  =  AB,  Ic  =  BC,  cd  =  CD,  etc.,  since  they 
are  parallels  included  between  parallels  (138)- 

Again,  the  corresponding  angles  of  these 
polygons  are  equal,  that  is,  a  =  A,  J  =  B,  c  =  C, 
etc.,  since  their  sides  are  parallel  and  lie  in  the 
same  direction  (492)- 

Therefore  the  polygons  ABODE  and  abcde  are 
mutually  equilateral  and  equiangular ;  that  is,  they  are 


pJiiS3£S.  249 

569.  Corollary. — Any  plane  section  of  a  prism,  paral- 
lel to  its  base,  is  equal  to  the  base ;  and  all  right  sections 
are  equal. 


PROPOSITION    II. 

570.  Theorem. — //  two  prisms  have  equivalent  bases, 
any  plane  sections  parallel  to  the  bases  are  equivalent. 

Demonstration. 

Let  M  and  N  be  any  two  prisms  having  equivalent  bases  B  and  B'; 
and  let  P  and  Q  be  sections  parallel  thereto. 

Then,  by  the  preceding  proposition, 

P  =  B, 
and  Q  =  B'  =  B , 

whence,  P  =  Q.        Q.  b.d. 


PROPOSITION    III. 

571.  Theorem. — //  three  faeces  including  a  triedral 
of  one  prism — complete  or  truncated — are  equal  respective- 
ly to  three  faces  including  a  triedral  of  the  other,  and 
similarly  placed,  the  prisms  are  equal. 

Demonstration. 

In  the  prisms  ^d  and  ^'d'  (Fig.  264),  let  ABCDE  equal  A'B'C'D'E, 
^Bba  =  ^'B'b'a',  and  BCcb  =  B'CV6'. 

Then  are  the  prisms  equal. 

For,  since  the  facial  angles  of  the  triedrals  B  and  B'  are  equal,  the 
triedrals  are  equal  (544),  and  being  applied  they  will  coincide. 

Now,  conceiving  ^'d'  as  applied  to  A^,  with  B'  in  B,  since  the  bases 
are  equal  polygons,  they  will  coincide  throughout ;  and  for  like  reason 
aB  will  coincide  with  a'B',  and  cB  witli  c'B'. 


250 


ELEMENTARY    GEOMETRY, 


Furthermore,  since  the 
coincide,  CD'  falls  in  CD,  and  as 
C'c'  falls  in  Cc,  and  D'd'  is  parallel 
to  CV,  and  Ud  to  Cc  (?),  [y'd'  falls 
in  Dd. 

In  like  manner,  £'e'  can  be 
shown  to  fall  in  Ee. 

Finally,  since  the  upper  bases 
have  the  angles  a'l'c'  and  ahc  co- 
incident, they  coincide  (444). 

Hence  the  prisms  can  be  super- 
imposed, and  are  therefore  equal. 


Fig.  264. 


Q.  E.  D. 


572.  Corollary. — Two  right  prisms  having  equal  bases 
and  equal  altitudes  are  equal. 

If  the  faces  are  not  similarly  arranged,  as  the  edges  are  perpendicular 
to  the  bases,  one  prism  can  be  inverted  and  then  superimposed  on  the 
other. 


PROPOSITION    IV. 

573.  Theorem. — Any  oblique  prism  is  equivalent  to  a 
right  prism,  whose  bases  are  right  sections  of  the  oblique 
prism,  and  whose  edge  is  equal  to  the  edge  of  the  oblique 
prism. 

Demonstration. 

Let  LB  be  an  oblique  prism,  of  which  abode 
an6ff//iil  are  right  sections,  and  gb  =  GB. 

Then  is  lb  equivalent  to  LB. 

For  the  truncated  prisms  ^G  and  eB  have  the 
faces  including  any  two  corresponding  triedrals,  as 
G  and  B,  respectively,  equal  and  similarly  placed 
(?),  whence  these  prisms  are  equal  (571)- 

Now,  from  the  whole  figure  take  away  prism 
ZG,  and  there  remains  the  oblique  prism  LB ;  also, 
from  the  whole  take  away  the  prism  cB,  and  there 
remains  the  right  prism  lb. 

Therefore,  the  right  prism  lb  is  equivalent  to 
the  oblique  prism  LB.     q.  e.  d. 


Fig.  265. 


PRISMS, 


251 


PROPOSITION    V. 

574.  Theorem. — The  opposite  faces  of  a  parcdlelopiped 
are  equal  and  parallel. 

Demonstration. 
Let  Ac  be  a  parallelopiped,  AC  and  ac  being  its  equal  bases  (560) 

Then  are  its  opposite  faces  equal  and  parallel. 

Since  the  bases  are  parallelograms,  AB  is  equal 
and  parallel  to  DC ;  and,  since  the  faces  are  paral- 
lelograms, ah,  is  equal  and  parallel  to  d[).     Hence, 

angle  aAB  =  d[>0, 

and  their  planes  are  parallel,  since  their  sides  are 
parallel  and  extend  in  the  same  directions. 

Therefore,  aS  and  dC  are  equal  (322)  and  parallel  parallelograms. 
In  like  manner  it  may  be  shown  that  aO  is  equal  and  parallel  to  M). 

Q.  E.  D. 


Fig.  266. 


PROPOSITION    VI 


575.  Theorem.— me 

bisect  each  other. 


diagonals  of  a  parallelopiped 


«. ^ 


AVw--^f 


Demonstration. 

Let  ABCD-A  be  a  parallelopiped  whose  diagonals  are  frD,  f/B,  cA, 
and  aO. 

Then  do  ftD,  <iB,  cA,  and  aC  bisect  each  other. 

Pass  a  plane  through  two  opposite  edges,  as 
&B  and  ^D. 

Since  the  bases  are  parallel  (?),  hd  and  BD  will 
be  parallel  (488),  and  &BD<Z  will  be  a  parallelo- 
gram.    Hence,  ftD  and  d^  are  bisected  at  o  (?). 

For  a  like  reason,  pnssing  a  plane  through  dc 
and  AB,  we  may  show  that  <iB  and  cA  bisect  each 
other,  and  hence  that  cA  passes  through  the  com- 
mon centre  of  c?B  and  &D. 

So  also  aO  is  bisected  by  5D,  as  appears  from 
passing  a  plane  through  ab  and  DC. 

Hence,  all  the  diagonals  are  bisected  at  o.     q.  b.  d. 


Fig.  267. 


252  ELE3IENTARY    GEOMETRY. 

576.    Corollary. — The  diagonals  of  a  rectangular  par- 
allelopiped  are  equal. 


PROPOSITION    VII. 

577.  Theorem.— ITz/e  diagonal  of  a  right  parallelo- 
piped  is  equal  to  the  square  root  of  the  sum  of  the  squares 
of  the  three  adjacent  edges  of  the  parallelepiped. 

Demonstration. 

Let  a,  b,  c  be  the  three  adjacent  edges  of  a  right  parallelepiped,  d 
the  diagonal  of  the  face  whose  edges  are  h  and  c,  and  D  the  diagonal  of 
the  parallelopiped. 

Then  ^»  =  &«  +  C  (?), 

and  D'  =  0?  -\-d'  =  a'  +  ¥  -vc^  (?), 


or  D  =  ^a?  +  &'  +  c^     q.  e.  d. 

578.  Corollary. — The  diagonal  of  a  cube  is  Vs  times 
its  edge. 


PROPOSITION    VIII. 

579.  Theorem.— ^e  area  of  the  lateral  surface  of  a 
right  prism  is  equal  to  the  product  of  its  altitude  into  the 
perimeter  of  its  base. 

Demonstration. 

The  lateral  faces  are  all  rectangles,  having  for  their  common  altitude 
the  altitude  of  the  prism  (563).  Whence  the  area  of  any  face  is  the 
product  of  the  altitude  into  the  side  of  the  base  which  forms  its  base ; 
and  the  sum  of  the  areas  of  the  faces  is  the  common  altitude  into  the  sum 
of  the  bases  of  the  faces,  that  is,  into  the  perimeter  of  the  base  of  the 
prism.     Ci.  '^.  d. 


CYLINDERS. 


253 


580.  A  Cyliiitlrical  Siu'face  is  a  surface  traced  by  a 
straight  line  moving  so  as  to  remain  constantly  parallel  to  its 
first  position,  while  any  point  in  it  traces 
some  curve.  The  moving  line  is  called  the 
Generate ixy  and  the  curve  traced  by  a  point 
of  the  line  the  Directrix. 

Illustration.—  Suppose  a  line  to  start  from 
the  position  AB,  and  move  towards  N  in  such  a 
manner  as  to  remain  all  the  time  parallel  to  its 
first  position  AB,  while  A  traces  the  curve 

A123456....M. 

The  surface  thus  traced  is  a  Cylindrical  Surface ; 
AB  is  the  Geneiatrix^  and  the  curve  ANM  the 
Directrix. 

Fig.  268. 

581.  A  Circular  Cylinder,  called  also  a  Cylinder  of 

Revolution,  is  a  solid  generated  by  the  revolution  of  a  rectan- 
gle around  one  of  its  sides  as  an  axis. 

Illustration. — Let  COAB  be  a  rectangle, 
and  conceive  it  revolved  about  CO  as  an  axis, 
taking  successively  the  positions  COAB', 
COA'B',  etc.;  the  solid  generated  is  a  Circular 
Oylinder,  or  a  cylinder  of  revolution.  The  re- 
volving side  AB  is  the  generatrix  of  the  surface, 
and  the  circumference  A  A' A"  (or  BB'B")  is  the 
directrix.  This  is  the  only  cylinder  treated  in 
Elementary  Geometry,  and  is  usually  meant 
when  the  word  Cijlinder  is  used  without  specify- 
ing the  kind  of  cylinder.  '"•a-  269. 

682.  The  Axis  of  the  cylinder  is  the  fixed  side  of  the  rectan- 
gle. The  side  of  the  rectangle  opposite  the  axis  generates  the 
Convex  Surface ;  while  the  other  sides  of  the  rectangle,  as 
OA  and  CB,  generate  the  Bases,  which  in  the  cylinder  of  revo- 
lution are  circles.  Any  line  of  the  surface  corresponding  to  some 
position  of  the  generatrix  is  called  an  Element  of  the  surface. 

583.  Any  section  of  a  cylinder  of  revolution  made  by  a  plane 
parallel  to  its  base  is  equal  to  its  base,  since  such  a  section  would 
be  a  circle  with  a  radius  equal  to  OA. 


254 


ELEMENTARY    GEOMETRY, 


584.  A  Right  Cylinder  is  one  whose  elements  are  perpen- 
dicular to  its  base.  In  such  a  cylinder  any  element  is  equal  to 
the  axis.     A  Cylinder  of  Revolution  (581)  is  right. 

685.  A  prism  is  said  to  be  inscribed  in  a  cylinder,  when  the 
bases  of  the  prism  are  inscribed  in  the  bases  of  the  cylinder,  and 
the  edges  of  the  prism  coincide  with  elements  of  the  cylinder. 


PROPOSITION    IX. 

586.  Theorem. — The  area  of  the  convex  surface  of  a 
cylinder  of  revolution  is  equal  to  the  product  of  its  axis 
into  the  circumference  of  its  base,  i.  e.,  277 RH,  H  being  the 
axis  and  R  the  radius  of  the  base. 

Demonstration. 

Let  AD  be  a  cylinder  of  revolution,  whose  axis  HO  =  H^  and  the 
radius  of  whose  base  is  OB  =  It, 

Then  is  the  area  of  its  convex  surface  ZnliH. 

Let  a  right  prism,  with  any  regular  polygon  for 
its  base,  be  inscribed  in  the  cylinder,  as  h-ahcdef. 

The  area  of  the  lateral  surface  of  the  prism  is 
HO  (=  Kb)  into  the  perimeter  of  its  base,  i.  e., 

Y\Oy.(ah^'l)c-\-cd  +  de  +  ef-\-fa). 

Now,  bisect  the  arcs  a5,  &c,  etc.,  and  inscribe  a 
regular  polygon  of  twice  the  number  of  sides  of  the 
preceding,  and  on  this  polygon  as  a  base  construct 
the  right  inscribed  prism  with  double  the  number  of 
faces  that  the  first  had.  The  area  of  the  lateral  sur- 
face of  this  prism  is 

HO  X  the  perimeter  of  its  base. 

In  like  manner,  conceive  the  operation  of  inscribing  right  prisms  with 
regular  polygonal  bases  continually  repeated ;  it  will  always  be  true  that 
the  area  of  the  lateral  surface  is  equal  to 

HO  X  the  perimeter  of  the  base. 


Fig.   270. 


CYLINDERS.  256 

By  continually  increasing  the  number  of  the  sides  of  the  inscribed 
polygon  in  this  manner,  the  perimeter  of  the  polygon  may  be  made  to 
differ  from  the  circumference  by  less  than  any  assignable  quantity,  i.  e., 
by  an  infinitesimal,  which  is  therefore  0  in  comparison  with  the  perimeter 
(341),  and  the  prism  of  an  infinite  number  of  faces  is  to  be  considered  as 
the  cylinder. 

Therefore,  the  area  of  the  convex  surface  of  the  cylinder  is  HO  into 
the  circumference  of  the  base. 

Finally,  if  E  is  the  radius  of  the  base,  27ri2  is  its  circumference.  This 
multiplied  by  H,  the  altitude,  i.  e.,  Hx  %-R,  or  'inRH,  is  the  area  of  the 
convex  surface  of  the  cylinder,     q.  e.  d. 


PROPOSITION    X. 

587.  Theorem. — Rectangular  parallelopipeds  are  to 
eoAih  other*  as  the  products  of  any  three  adjacent  edges. 

Demon^stration. 

Let  the  adjacent  edges  of  one  rectangular  parallelopiped,  P,  be  three 
lines,  which  we  will  call  A,  B,  and  C^  and  of  another,  Q,  the  three  lines 
a,  bf  c. 

Then  ^  =  ^^^4^- 

Q         axoxc 

For  Aj  jP,  C,  a,  5,  and  c  are  at  least  commensurable  by  an  infinitesimal 
unit.  Let  the  common  measure  of  the  edges  be  i ;  and  let  it  be  contained 
in  A  m  times,  in  ^  n  times,  in  C p  times,  in  a  ^  times,  in  5  r  times,  and  in 
e  8  times,  so  that 

ABC 

a  ^  J  c 

g  =  T  ,      -       r  =  T  ,    and    «  =  t  • 

Now  let  A  and  5  be  the  sides  of  the  rectangular  base  of  P,  and  G  its 
altitude,  and  a,  5,  and  c  corresponding  edges  of  Q.  The  base  of  P  may  be 
conceived  as  divided  into  mn  units  of  surface.  If  upon  each  of  these  we 
conceive  a  cube  described,  there  will  be  mn  such  cubes.  Now,  of  these 
layers  of  cubes  there  will  be  p  in  the  entire  parallelopiped  P.  Hence  P 
will  be  composed  of  mnp  equal  cubes.     In  like  manner,  Q  may  be  shown 

*  This  means  that  their  volumes  are  to  each  other. 


256  ELEMENTARY    GEOMETRY, 

to  be  composed  of  qr*  equal  cubes,  each  equal  to  one  of  the  mnp  cubes 
which  compose  P. 

P      mnp 

and  substituting  their  values  for  w,  /i,  2*1  ?,  ^%  and  s,  we  have 

-       -  X  - 
P        t        i»        AxBxG  „ 

^  =  -^ i: = J. Q-  E-  J^- 

O        a     b      c  axo  X.C 

V     X    T     X     T 
I  I  t 


PROPOSITION    XI. 

588.  Theorem. — The  volume  of  a  rectangular  paral- 
lelopiped  is  equal  to  the  product  of  its  three  adjacent 
edges. 

Demonstration. 

Let  P  be  any  rectangular  parallelopiped  whose  adjacent  edges  are 
Af  Bf  and  C,  and  let  Q  be  the  proposed  unit  of  measure,  whose  edges 
are  each  I. 

Then,  by  the  last  proposition, 

P  _  Aj<B^G 
Q  ""    1x1x1  ' 

or,  P  z=  {AxByC)yQ 

Thus,  P  contains  the  unit  Q  AxBx  C  times.  Hence,  AxBx  G  is 
the  volume  of  P.    q.  e.  d. 

589.  Corollary  1. — The  volume  of  a  cube  is  the  third 
power  of  its  edge, 

590.  Scholium. — This  fact  gives  rise  to  the  term  cube,  as  used  in 
arithmetic  and  algebra,  for  "  third  power.'* 

591.  Corollary  2. — The  volum^e  of  a  rectangular  par- 
allelopiped is  equal  to  the  product  of  its  altitude  into  the 
area  of  its  base,  the  linear  unit  being  the  same  for  the 
measure  of  all  its  edges. 

*  For  other  demonstrations  see  Appendix. 


CYLINDERS,  ^67 


PROPOSITION    XII. 

692.  Theorem. — Tlie  volume  of  amj  prism,  or  of  any 
solid  whose  plane  sections  parallel  to  the  base  are  all  equal 
to  the  base,  is  equal  to  that  of  a  rectangular  pardUelopipecl 
having  an  equivalent  base  and  the  same  altitude,  and 
hence  is  equal  to  the  product  of  its  base  into  its  altitude. 

Demonstration. 

Let  Q  be  any  prism  or  solid  whose  plane  sections  parallel  to  its  base 
are  equal  to  its  base,  and  P  a  rectangular  parallelepiped  of  the  same 
altitude,  and  whose  base  li  =  B',  the  base  of  the  first  solid. 

Then  is  volume  Q  =  volume  P. 

If  Q  be  a  prism,  any  plane  section  parallel  to  its  base  is  equal  to  its 
base  (?) ;  hence  the  case  is  the  same  whether  Q  be  a  prism  or  any  other 
solid  having  its  plane  sections  parallel  to  its  base  equal  to  its  base. 

Now  conceive  two  planes  to  start  from  coincidence  with  B  and  B*  at 
the  same  time,  and  move  upward  at  the  same  rate,  generating  the  solids 
P  and  Q.  As  these  sections  are  always  equivalent  to  each  other,  since 
each  is  constantly  equal  to  /?  or  5',  they  generate  equal  volumes  in  equal 
times,  and  by  reason  of  the  equal  altitudes  of  the  two  solids,  both 
volumes  are  generated  in  the  same  time.  Hence  the  two  volumes  are 
equivalent,     q.  e.  d. 

593.  Corollary  1. — The  volume  of  a  right  prism  is 
equal  to  the  product  of  its  edge  into  its  base. 

594.  Corollary  2.  — Prisms  of  the  same  altitude  are  to 
each  other  as  their  bases  ;  and  prisms  of  the  sam^e  or  equiv- 
alent bases  are  to  each  other  a§  their  altitudes ;  and,  in 
general,  prisrns  are  to  each  other  as  the  products  of  their 
bases  and  altitudes. 


258  ELEMJEIiTAMY    GEOMETRY. 


PROPOSITION     XIII. 

696.  Theorem. — The  volume  of  a  cylinder  of  revolu- 
tion is  equal  to  the  product  of  its  base  and  altitude,  i.  e., 
ttR^H,  H  being  the  altitude  and  R  the  radius  of  the  base. 

Demonstration. 

By  (592)  the  volume  of  sucli  a  cylinder  is  equal  to  the  product  of  its 
base  into  its  altitude,  since  all  plane  sections  parallel  to  its  base  are  equal 
thereto  (583). 

But  the  base  is  a  circle  whose  radius  is  i2,  the  area  of  which  is  ttR^  (?). 
Hence  the  volume  of  the  cylinder  is  ^x  TTJS^  or  TrijJ^fi;     q.  e.  d. 

596.  Corollary. — The  volume  of  any  cylinder  is  equal 
to  the  product  of  its  base  into  its  altitude. 

This  can  be  demonstrated  in  a  manner  altogether  analogous  to  the 
case  given  in  the  proposition. 


697.  Similar  Solids  are  such  as  have  their  corresponding 
solid  angles  equal  and  their  homologous  edges  proportional. 

698.  Similar  Cylinders  of  revolution  are  such  as  have 
their  altitudes  in  the  same  ratio  as  the  radii  of  their  bases. 

699.  Homologous  Edges  of  similar  solids  are  such  as 
are  included  between  equal  plane  angles  in  corresponding  faces. 

Illustration. — The  idea  of  similarity  in  the  case  of  solids  is  the 
same  as  in  the  case  of  plane  figures,  viz.,  that  of  likeness  of  form.  Thus, 
one  would  not  think  such  a  cylinder  as  one  joint  of  stovepipe  similar  to 
another  composed  of  a  hundred  joints  of  the  same  pipe.  One  would  be 
long  and  very  slim  in  proportion  to  its  length,  while  the  other  would  not 
be  thought  of  as  slim.  But,  if  we  have  two  cylinders  the  radii  of  whose 
bases  are  2  and  4,  and  whose  lengths  are  respectively  6  and  13,  we  readily 
recognize  them  as  of  the  same  shape :  they  are  similar. 


PRISMS    AND     CYLINDERS. 


PROPOSITION    XIV. 

600.  Theorem, — The  altitudes  of  two  similar  prisms 
are  to  each  other  as  any  two  homologous  edges,  and  the 
areas  of  corresponding  faces  are  to  each  other  as  the 
squares  of  any  two  homologous  edges,  or  as  the  squares  of 
the  altitudes. 


Demonstration. 

Let  P  and  />  be  any  two  similar  prisms,  11  and  h  their  altitudes,  Ka 
and  k'a'  two  homologous  edges,  and  A6  and  Mb'  two  corresponding 
faces. 


Then  is 


H      Ha 

T-  =  jTi  >  or  as  aTiy  other  two  homologouB  edges : 


and 


Aft 

A'6' 


Aa 


H' 


>  /  ,i=  -^  '     *•  *M   as  the   squares    of   any 
A' a'*       h^  n  .7 


other  two  homologous  edges,  or  as  the  squares  of  the  altitudes. 

From  the  homologous  vertices  a  and  a'  let  fall  the  perpendiculars  a\ 
and  a' I',  and  draw  Al  and  AT. 

a\  =  H,m(\a'\'  =  hO). 

Now,  since  the  prisms  are 
similar,  they  may  be  so  placed 
that  their  homologous  edges 
wHl  be  parallel ;  hence,  let  AB 
be  parallel  to  A'B',  AE  to  A'E', 
and  ^A  to  a'A'.  Then  is  al 
parallel  to  a'!',  and  Al  to  AT, 
and  the  triangles  aAl  and  a' AT 
are  similar. 

Whence  we  have 


H 
h 


Aa 

k^a" 


Fig.  271. 


or  as  any  other  two  homolo- 
gous edges,  since  by  definition 
any  two  homologous  edges  bear  the  same  ratio.    Q.  e.  d. 


260 


ELEMENTARY    GEOMETRY, 


Again,  since  the  corre- 
sponding faces  Kb  and  A'^' 
have  their  homologous  sides 
proportional  (597),  and  their 
homologous  angles,  as  aAB  and 
a'A'B',  equal,  being  the  ho- 
mologous facial  angles  of  equal 
triedrals,  the  faces  are  similar 
plane  figures,  and 


ka 


IP 


or  as  the  squares  of  any  two 
homologous  edges,     q.  b.  d. 


Fig.  271. 


601.  Corollary. — Tlie  corresponding  faces  of  any  two 
similar  solids  are  to  each  other  as  the  squares  of  any  two 
homologous  edges  of  the  solids. 


PROPOSITION    XV. 

602.  Theorem. — The  lateral  surfaces  of  similar  prisms 
are  to  each  other  as  the  squares  of  any  two  homologous 
edges,  or  as  the  squares  of  the  altitudes  of  the  prisms. 

Demonstration. 

Let  Af  B,  C,  I>,  etc.,  and  a,  b,  c,  d,  etc.,  be  the  corresponding 
faces  of  two  similar  prisms,  and  M  and  in  any  two  homologous  edges, 
and  H  and  h  the  altitudes. 


By  the  last  proposition, 


A 

a  " 

Jf2 

B      M' 

CM'          D 

ce, 

A_B_G_ 
a  ~  b       c 

—  ,  etc.  =  -—^ , 
d               VI- 

M^ 


etc. 


PRISMS    AND    CYLINDERS.  261 

and,  by  coiuposition, 

a  +  6  +  c  +  rf,  etc.     ~  m^  ~  h^  ^^'     ^•^^^■ 

603.  Corollary. — The  entire  surfaces  of  any  two  simi- 
lar solids  are  to  each  other  as  the  squares  of  any  two 
homologous  edges. 


PROPOSITION    XVI. 

604.  Theorem. — The  volumes  of  similar  prisms  are  to 
each  other  as  the  cubes  of  their  homologous  edges,  and  as 
the  cubes  of  their  altitudes. 

Demonstration. 

Let  V  and  v  be  the  volumes  of  any  two  similar  prisms,  M  and  m 
any  two  homologous  edges,  and  ^  and  h  their  altitudes. 

Then  is  -=-:  =  ^'- 

Let  B  and  h  be  the  bases  of  the  prisms ;  whence  their  volumes  are 
5xJy  and  hxh  respectively  (592). 


By  (600), 


B  ^M^^H^ 

b  ~  W  ~   A^ 


Bat  f=-=f(')- 

h         m        h    ^  ^ 


Multiplying,  -r-T-  ^  —  -  -^  ^  Ti'    Q.  E.  D. 


262  ELEMENTARY    GEOMETRY. 


PROPOSITION    XVII. 

605.  Theorem. — The  convex  surfaces  of  similar  cylin- 
ders of  revolution  are  to  each  other  as  the  squares  of  their 
altitudes,  and  as  the  squares  of  the  radii  of  their  bases. 

Demonstration. 

Let  H^rxA  h  be  the  altitudes,  and  R  and  r  the  radii  of  the  bases  of 
two  similar  cylinders. 

The  convex  surfaces  are  27ri2^and  ^tttK  (586). 

T^  2nRH      RH      R      H 

•Now,  — — -  =  — -  =  -  X  ^  . 

2irrA         rh         t       h 

By  hypothesis,  —  =  —  • 

Whence,  by  substitution,  we  have 

27rrh    ~  h?  ' 

,  2itRh     m 

and  — — ~  -  —•    Q.  E.  D. 

2nrh         T^ 


PROPOSITION    XVIII. 

606.  Theorem. — The  volumes  of  similar  cylinders  of 
revolution  are  to  each  other  as  the  cubes  of  their  altitudes, 
or  as  the  cubes  of  the  radii  of  their  bases. 

Demonstration. 

Let  H  and  fi  be  the  altitudes  of  two  similar  cylinders  of  revolution, 
M  and  r  the  radii  of  their  bases,  and  V  and  v  their  volumes. 

V       H^       R^ 


PRISMS    AND    CYLINDERS  263 


For,  by  (595),  V  =  itHR', 

and  V  =  nhr^. 

Hence,  —  =  — ir-r  —  ^tt 


and,  since 


-  =  -(?) 


we  have,  by  substitution,  —  =  —  =  -—.    q.  e.  D. 


607.  Scholium.— It  is  a  general  truth,  that  the  surfaces  of  similar 
solids,  of  any  form^  are  to  each  other  as  the  squares  of  homologous  lines; 
and  their  volumes  are  as  the  cubes  of  such  lines.  These  truths  will  be 
further  illustrated  in  the  following  section,  but  the  methods  of  demon- 
stration will  be  seen  to  be  the  same  as  used  in  this. 


EXERCISES. 

1.  A  farmer  has  two  grain  bins  which  are  parallelo- 
pipeds.  The  front  of  one  bin  is  a  rectangle  6  feet  long  by  4  high, 
and  the  front  of  the  other  a  rectangle  8  feet  long  by  4  high. 
They  are  built  between  parallel  walls  5  feet  apart.  The  bottom 
and  ends  of  the  first,  he  says,  are  "  square  "  (he  means,*  it  is  a 
rectangular  parallelopiped),  while  the  bottom  and  ends  of  the 
other  slope,  i.  e.,  are  oblique  to  the  front.  What  are  the  rela- 
tive capacities  of  the  bins  ? 

2.  How  many  square  feet  of  boards  in  the  walls  and  bottom 
of  the  first  bin  mentioned  in  Ex.  1  ? 

3.  An  average  sized  honey  bee's  cell  is  a  right  hexagonal 
prism,  .8  of  an  inch  long,  with  faces  ^  of  an  inch  wide.  The 
width  of  the  face  is  always  the  same,  but  the  length  of  the  cell 
varies  according  to  the  space  the  bee  has  to  fill.  Are  honey  bees* 
cells  similar  ?  Is  a  honey  bee's  cell,  of  the  dimensions  given 
above,  similar  to  a  wasp's  cell,  which  is  1.6  inches  long,  and 
whose  face  is  .3  of  an  incli  wide?  What  are  the  relative  capaci- 
ties of  the  wasp's  cell  and  the  honey  bee's? 


264  ELEMENTARY     GEOMETRY. 

4.  How  many  square  inches  of  sheet  iron  docs  it  take  to  make 
a  joint  of  7-inch  stovepipe  2  feet  4  inches  long,  allowing  an  inch 
and  a  half  for  making  the  seam  ? 

5.  A  certain  water-pipe  is  3  inches  in  diameter.  How  much 
water  is  discharged  through  it  in  24  hours,  if  the  current  flows 
3  feet  per  minute  ?  How  much  through  a  pipe  of  twice  as  great 
diameter,  at  the  same  rate  of  flow  ? 

6.  What  is  the  ratio  of  the  length  of  a  hogshead  holding  125 
gallons,  to  the  length  of  a  keg  of  the  same  shape,  holding 
8  gallons  ? 

7.  What  are  the  relative  amounts  of  cloth  required  to  clothe 
three  men  of  the  same  form  (similar  solids),  one  being  5  feet 
high,  another  5  feet  9  inches,  and  tlic  other  6  feet,  provided  they 
dress  in  the  same  style?  If  the  second  of  these  men  weighs 
156  lb.,  what  do  the  others  weigh ? 

8.  If  a  man  5 J  feet  high  weighs  160  lb.,  and  a  man  3  inches 
taller  weighs  180  lb.,  which  is  the  stouter  in  proportion  to  his 
height  ? 

9.  I  have  a  prismatic  piece  of  timber,  from  which  I  cut  two 
blocks,  both  5  feet  long  measured  along  one  edge  of  the  stick ; 
but  one  block  is  made  by  cutting  the  stick  square  across  (a  right 
section),  and  the  other  by  cutting  both  ends  of  it  obliquely, 
making  an  angle  of  45°  with  the  same  face  of  the  timber.  Which 
block  is  the  greater  ?     Which  has  the  greater  lateral  surface  ? 

10.  How  many  cubic  feet  in  a  log  12  feet  long  and  2  feet  and 

5  inches  in  diameter  ?  How  many  square  feet  of  inch  boards 
can  be  cut  from  such  a  log,  allowing  one-quarter  for  waste  in 
slabs  and  sawing  ? 

11.  How  many  square  feet  of  sheet  copper  will  it  take  to  line 
the  sides  and  bottom  of  a  cylindrical  vat  (cylinder  of  revolution) 

6  feet  deep,  if  the  diameter  of  the  bottom  is  4  feet?  How  many 
barrels  does  such  a  vat  contain  ? 

12.  What  are  the  relative  capacities  <  -  of  rovolntion 
of  the  same  diameter,  but  of  different  len..  \'5..  <  of  those 
of  the  same  length,  but  of  different  diamuttrs 


PYRA 


W    CONES, 


265 


^irCTiOH  lY, 


OF    PYRAMIDS    AND    CONES. 

609.  A  Pyramid  is  a  solid  having  a  polygon  for  its  base, 
and  triangles  for  its  lateral  faces.  If  the  base  is  also  a  triangle, 
it  is  called  a  triangular  pyramid,  or  a  tetraedron  (i.  e.,  a  solid  with 
four  faces).  The  vertex  of  tlie  polyedral  angle  formed  by  the 
lateral  faces  is  the  Vertex  of  the  pyramid. 

610.  The  Altitude  of  a  pyramid  is  the  perpendicular  dis- 
tance from  its  vertex  to  the  plane  of  its  base. 

611.  A  Right  Pyramid  is  a  pyramid  whose  base  is  a  regu- 
lar polygon,  and  the  perpendicular  from  whose  vertex  falls  at 
the  centre  of  the  base.     This  perpendicular  is  called  the  axis, 

612.  A  Frustum  of  a  pyramid  is  a  portion  of  the  pyramid 
intercepted  between  the  base  and  a  plane  parallel  to  the  base. 
If  the  cutting  plane  is  not  parallel  to  the  base,  the  portion  inter- 
cepted is  called  a  Truncated  pyramid. 

613.  The  Slaut  Height  of  a  right  pyramid  is  the  altitude 
of  one  of  the  triangles  which  form  its  faces.  The  Slant  Height 
of  a  Frustum  of  a  right  pyramid  is  the  portion  of  the  slant  height 
of  the  pyramid  intercepted  between  the  bases  of  the  frustum. 


Fig.  272. 

Illustrations.— The  student  will  be  able  to  find  illustrations  of  the 
definitions  in  the  above  figures. 


266  ELEMENTARY    GEOMETRY. 

614.  A  Conical  Surface  is  a  surface  traced  by  a  line 
which  passes  througli  a  fixed  point,  while  any  other  point  traces 
a  curve.  The  line  is  the  Generatrix,  and  the  curve  the  Direc- 
trix. The  fixed  point  is  the  Vertex.  Any  line  of  the  surface 
corresponding  to  some  position  of  the  generatrix  is  called  an 
Element  of  the  surface. 

615.  A  Cone  of  Revolution  is  a  solid  generated  by  the 
revolution  of  a  right-angled  triangle  around  one  of  its  sides, 
called  the  Axis.  The  hypotenuse  describes  the  Convex  Surface 
of  the  cone,  and  corresponds  to  the  generatrix  in  the  preceding 
definition.  The  other  side  of  tlie  triangle  describes  the  Base. 
This  cone  is  right,  since  the  perpendicular  (the  axis)  falls  at  the 
centre  of  the  base.  The  Slayit  Height  is  the  distance  from  the 
vertex  to  tlie  circumference  of  the  base,  and  is  the  same  as  the 
hypotenuse  of  the  generating  triangle. 

616.  The  terms  Frustum  and  Truncated  are  applied  to 
the  cone  in  the  same  manner  as  to  the  pyramid. 

617.  A  pyramid  is  said  to  be  Inscribed  in  a  cone  when  the 
base  of  the  pyramid  is  inscribed  in  the  base  of  the  cone,  and  the 
edges  of  the  pyramid  are  elements  of  the  surface  of  the  cone. 
The  two  solids  have  a  common  vertex  and  a  common  altitude. 

618.  If  the  generatrix  be  considered  as  an  indefinite  straight 
line  passing  through  a  fixed  point,  the  portions  of  the  line  on 
opposite  sides  of  the  point  will  each  describe  a  conical  surface. 
These  two  surfaces,  which  in  general  discussions  are  considered 
but  one,  are  called  Nappes.  The  two  nappes  of  the  same  cone 
are  evidently  alike. 

Illustration. —In  Fig,  273,  (a)  represents  a  conical  surface  which 
has  the  curve  ABC  for  its  directrix^  and  SA  for  its  generatrix.  The  nu- 
merals indicate  the  successive  positions  of  the  point  A,  as  it  passes  around 
the  curve,  while  the  point  S  remains  fixed,  (b)  represents  a  Gorie  of  Rev- 
olution^ or  a  right  cone  with  a  circular  base.  It  may  be  considered  as 
generated  in  the  general  way,  or  by  the  right-angled  triangle  SOA  revolv- 
ing about  SO  as  an  axis.     SA  describes  the  convex  surface,  and  OA  the 


PYRAMIDS    AND    CONES. 


267 


Fig.  273. 

base.  The  figure  (c)  represents  the  Frustum  of  a  cone,  the  portion  above 
the  plane  abc  being  supposed  removed.  Figure  {d)  represents  the  two 
Nappes  of  an  oblique  cone. 


PROPOSITION    I. 

619.  Theorem. — Any  section  of  a  pyramid  made  by  a 
plane  parallel  to  its  base  is  a  polygon  similar  to  the  base. 

Demonstratiox. 

Let  ahcde  be  a  section  of  the  pyramid 
S-ABCDE  made  by  a  plane  parallel  to  ABODE. 

Then  is  dbaJe  similar  to  ABODE. 

Since  AB  and  ab  are  intersections  of  two 
parallel  planes  by  a  third  plane,  they  are  paral- 
lel (0.  So  also  Ic  is  parallel  to  BO,  cd  to  OD, 
etc.  Hence,  angle  5  =  B,  c  =  0,  etc.  (?),  and  the 
polygons  are  mutually  equiangular.     Again, 

ab       Sb  ,      be        Sb 


AB-SB-    ""^     Be  =  is<^)- 


Fig.  274. 


Hence 


(?)• 


c^  _   bc^  a5  _  AB 

AB  ~  BO '     ^^    6c  ~  BC 
In  like  manner,  we  can  show  that 

bc_ 

ed~  CD' 

Therefore,  dhcde  and  ABODE   are   mutually  equiangular,  and  have 
tlu'ir  corresponding  *tles  proportional,  and  are  consequently  similar. 

Q.  E.  D. 


etc. 


268 


ELEMENTARY    GEOMETRY. 


PROPOSITION    II. 

620.  Theorem. — //  two  pyramids  of  equal  altitudes 
are  cab  by  planes  equally  distant  from  and  parallel  to 
their  bases,  the  sections  are  to  each  other  as  the  bases. 


Demonstration^.  ^ 

Let  S-ABC  and  S'-A'B'C'D'E'  be  two  pyramids  of  the  same  altitude, 
cut  by  the  planes  abc  and  a'b'dd'e' ,  parallel  to  and  at  equal  distances 
from  their  bases. 


Then  is 


abc 


ABC 


< 


a'h'c'd'i 


ABODE' 


For,  conceive  the  bases  in  the 
same  plane.  Let  SP  and  S'P'  be  the 
equal  altitudes,  and  Sp  =  S'p'  the 
distances  of  the  cutting  planes  from 
the  vertices. 

Conceive  a  plane  passing  through 

the  vertices  parallel  to  the  plane  of 

the  bases.     This  plane,  together  with 

the  plane  in  which  the  sections  lie, 

and  that  in  which  the  bases  lie,  make 

three  parallel  planes  which  cut  the 

lines  SA,  SB,  S'A',  S'B',  SP,  and  S'P', 

whence 

SB       SP  _  S^B 

S6'  "  Sp~  S'V 


g.   275. 


ST' 

S'p' 


Also,  since  the  planes  ASB  and  A'S'B'  are  cut  by  parallel  planes  in 
AB,  (il,  A'B',  and  a'V,  ab  is  parallel  to  AB,  and  a'h'  to  A'B' ;  whence. 


Now 


and 


AB       SB 

ab  ~  S6 ' 

A'B'       SB' 

ABC 

abc    ~ 

^'  ^'^  -  S^''  ^  ^' 

A'B'C'D'E 

a'h'c'd'e' 

:'      AM3''      s^' 

PYRAMIDS    AND    CONES. 
Hence,  by  equality  of  ratios, 


269 


ABC       A'B'C'D'E'  ahc 

— r-  =  — ,.,  ,,,-r,     or 

aJbc  ah'cde 


ABC 


a'h'c'd'ef       A'B'C'D'E' 


(?).      Q.  E.  D. 


621.  Corollary. — //  two  pTjramids  having  equivalent 
bases  and  equal  altitudes  are  cut  by  planes  parallel  to  and 
equidistant  from  their  bases,  the  sections  are  equivalent. 


PROPOSITION   III. 

622.  Theorem. — The  area  of  the  lateral  surface  of  a 
right  pyramid  is  equal  to  the  perimeter  of  the  base  multi- 
plied bij  one-half  the  slant  height. 


Demonstration. 

The  faces  of  such  a  pyramid  are  equal  isosceles  tri- 
angles (?),  whose  common  altitude  is  the  slant  height  of 
the  pyramid  (?). 

Hence,  the  area  of  these  triangles  is  the  product  of 
one-half  the  slant  height  into  the  sum  of  their  bases.  But 
this  sum  is  the  perimeter  of  the  base. 

Hence  the  area  is  equal  to  the  perimeter  of  the  base 
multiplied  by  one-half  the  slant  height,     q.  e.  d. 


Fig.  276. 


623.  Corollary. — The  area  of  the  lateral 
surface  of  the  frustum  of  a  right  pyrajnid  is 
equal  to  the  product  of  its  slant  height  into 
half  the  sum,  of  the  perimeters  of  its  bases. 

The  proof  is  based  upon  (350)  and  definitions. 


/<w^ 


Fig.  277. 


270  ELEMEi^TARY    GEOMETRY, 


PROPOSITION    IV. 

624.  Theorem. — The  area  of  the  convex  surface  of  a 
cone  of  revolution  (a  j^ight  cone  with  a  circular  base)  is 
equal  to  the  product  of  the  circumference  of  its  base  and 
one-half  its  slant  height,  i.  e.,  -nRH',  R  being  the  radius  of 
the  base,  and  H'  the  slant  height. 

Demonstration^. 

In  the  circle  which  forms  the  base  of  the  cone, 
conceive  a  regular  polygon  inscribed,  as  abcdef.  Join- 
ing the  vertices  of  the  angles  of  this  polygon  with  the 
vertex  of  the  cone,  there  will  be  constructed  a  right 
pyramid  inscribed  in  the  cone.  Now,  if  the  arcs  sub- 
tended by  the  sides  of  this  polygon  be  bisected,  and 
these  are  again  bisected,  etc.,  and  at  every  step  a  right 
pyramid  is  conceived  as  inscribed,  it  will  always  remain 
true  that  the  lateral  surface  of  the  pyramid  is  the  pe- 
rimeter of  its  base  into  half  its  slant  height. 

But,  as  the  number  of  faces  of  the  pyramid  is  in-  '^' 

creased,  the  perimeter  of  the  base  approaches  the  circumference  of  the 
base  of  the  cone  as  its  limit,  and  hence  the  slant  height  of  the  pyramid 
approaches  the  slant  height  of  the  cone,  and  the  lateral  surface  of  the 
pyramid  approaches  the  convex  surface  of  the  cone  as  their  limits,  and  all 
reach  their  limits  simultaneously. 

Therefore,  at  the  limit  we  still  have  the  same  expression  for  the  area 
of  the  convex  surface,  that  is,  the  circumference  of  the  base  multiplied 
by  half  the  slant  height. 

Finally,  if  E  is  the  radius  of  the  base,  its  circumference  is  2;r^,  and 
if  being  the  slant  height,  we  have  for  the  area  of  the  convex  surface 
2jR  X  ^H\  or  ttBH'.    q.  e.  d. 

625.  Corollary  1. — The  area  of  the  convex  surface  of  a 
cone  is  also  equal  to  the  product  of  the  slant  height  into 
the  circumference  of  the  circle  parallel  to  tJie  base,  and 
midway  between  the  base  and  vertex. 

This  follows  directly  from  the  fact  that  the  radius  of  the  circle  mid- 
way between  the  base  and  vertex  is  one-half  the  radius  of  the  base,  i.  e., 
^B  (?),  whence  its  circumference  is  tt/?.  Now,  tvRxII'  is  the  area  of  the 
convex  surface,  by  the  proposition. 


^IDS    AND    CONES. 


271 


LLAEY  '^.  —  rhe  area  of  the  convej&surfaceof 
the  fru  of  a  cone  is  equal  to  the  product  of  its  slant 

licight  r%to  half  the  sum  of  the  circumferences  of  its 
bases;  i  e.,  T{R-\-r)jr.  R  and  r  being  the  radii  of  its 
bases,  and  M'  its  slant  height. 

From  the  corresponding  property  of  the  frustum  of  a  pyramid,  the 
student  will  be  able  to  deduce  the  fact  that  ^  {2nB-\-%Tn')  H'  or  tt  {H-\-r)  IT 
is  the  area  of  this  surface  by  the  same  line  of  argument  used  in  the 
demonstration  of  the  main  theorem. 

627.  Corollary  3. — The  area  of  the  convex  surface  of 
the  frustum  of  a  cone  is  equal  to  the  product  of  its  slant 
lieight  into  the  circumference  of  the  circle  midway  between 
the  bases. 

The  radius  of  the  circle  midway  between  the  bases  is  ^  (/•+/?),  whence 
its  circumference  is  7r(r  +  i2).  Now,  Tr{r-\-R)xH'  is  the  area  of  the  con- 
vex surface  of  the  frustum,  by  the  preceding  corollary. 


PROPOSITION     V. 

628.  Theorem.  —  Two  pyram^ids  having  equivalent 
bases  and  the  same  altitudes  are  equivalent,  i.  e.,  equal  in 
volume. 

First  Demonstration. 

Let  S-ABCD  and  S'-A'B'C'D'E'  be  two  pyramids  having  the  same 
altitudes,  and  base  ABCD  equivalent  to  base  ABODE',  /.  e,,  equal  in 
area. 


Then  is  pyramid  S-ABCD 
equivalent  to  S'-A'B'C'D'E',  L  ^., 
equal  in  volume. 

For,  conceive  the  bases  to  be 
in  the  same  plane,  and  a  plane  to 
start  from  coincidence  with  the 
plane  of  the  bases,  and  move  to- 
ward the  vertices,  remaining  all 
the  time  parallel  to  the  bases. 


Fig.  279. 


2^2  ELEMENTARY    GEOMETRY. 

Now  each  of  the  sections  of  the  pyramids  made  by  this  plane  may  be 
conceived  as  a  varying  polygon  which  generates  its  respective  pyramid. 
And  as  these  polygons  are  always  equivalent,  and  move  at  tbe  same  rate, 
they  generate  equal  volumes  in  equal  times.  Moreover,  as  the  bases  of 
the  pyramids  are  in  the  same  plane,  and  their  altitudes  are  equal,  the 
polygons  generate  their  respective  pyramids  in  the  same  time.  Hence 
these  volumes  are  equal,     q.  e.  d. 

Second  Demonstkation. 

Consider  the  pyramids  divided  into  an  infinite  number  of  laminae  of 
equal  but  infinitesimal  thickness,  as  mc,  m'c\  parallel  to  the  bases.  Now 
each  lamina  in  one  will  have  a  corresponding  lamina  in  the  other  at  the 
same  distance  from  the  base  since  the  laminae  are  of  equal  thickness,  and 
hence  equivalent  to  it. 

Hence  the  pyramids  are  composed  of  an  equal  number  of  equivalent 
laminae,  and  are  consequently  equivalent,    q.  e.  d. 


PROPOSITION    VI. 

629.   Theorem. — The  volume  of  a  triangular  pyramid 
is  equal  to  one-third  the  product  of  its  base  and  altitude,  " 


>i 


Demonstration^. 


Let  S-ABC  be  a  triangular  pyramid,  whose  altitude  is  H, 

Then  is  the  volume  equal  to 
^H  X  area  ABC. 

For,  through  A  and  B  draw  Aa  and  B& 
parallel  to  SC;  and  through  S  draw  Sa 
and  S&  parallel  to  CA  and  CB,  and  join  a 
and  J);  then  Sa6-ABC  is  a  prism  with  its 
base  equal  to  the  base  of  the  pyramid. 

Now,  the  solid  added  to  the  given  pyr- 
amid is  a  quadrangular  pyramid  with  abB^ 
as  its  base,  and  its  vertex  at  S. 

Divide  this  into  two  triangular  pyra-  '^'9-  280. 

mids  by  drawing  aB  and  passing  a  plane  through  SB  and  aE.  These  tri- 
angular pyramids  are  equivalent,  since  they  have  equal  bases,  aAB  and 
abS,  and  a  common  altitude,  the  vertices  of  both  being  at  S. 


PYRAMIDS    AND    CONES. 


273 


Again,  S-«JB  may  be  considered  as  having  abS  (equal  to  ABC)  as  its 
base,  and  the  altitude  of  the  given  pyramid  (equal  to  the  altitude  of  the 
prism)  for  its  altitude,  and  hence  as  equivalent  to  the  given  pyramid. 

Thus  the  prism  KBCdbS  is  divided  into  the  three  equivalent  pyramidsi, 
S-ABC,  B-a&S,  and  S-«BA. 

Hence,  the  pyramid  S-ABC  is  one-third  the  prism  Sa5-ABC,  which 
has  the  same  base  and  altitude. 

But  the  volume  of  the  prism  is 

H  X  area  ABC. 
Therefore  the  volume  of  the  pyramid  S-ABC  is 

^H  X  area  ABC.     Q.  e.  d. 


630.  Corollary  1. — The  volume  of  any 
pyramid  is  equal  to  one-third  the  product 
of  its  base  and  altitude. 

Since  any  pyramid  can  be  divided  into  triangular 
pyramids  by  passing  planes  through  any  one  edge,  as 
SE,  and  each  of  the  other  edges  not  adjacent,  as  SB 
and  SC,  the  volume  of  the  pyramid  is  equal  to  the  sum 
of  the  volumes  of  several  triangular  pyramids  having 
the  same  altitude  as  the  given  pyramid,  and  the  sum 
of  whose  bases  is  the  base  of  the  given  pyramid. 


Fig.  281. 


631.  CoROLLART  2. — Pyramids  having  equivalent  bases 
are  to  each  other  as  their  altitudes ;  such  as  have  equal 
altitudes  are  to  each  other  as  their  bases  ;  and^  in  general, 
pyramids  are  to  each  other  as  the  products  of  their  bases 
and  altitudes. 

Exercise.— A  Regular  Tetraedron  is  a  triangular  pyr- 
amid whose  base  is  an  equilateral  triangle  and  each  of  whose 
lateral  faces  are  equal  to  the  base.  What  is  the  volume  of  such 
a  tetraedron  whose  edge  is  1  inch  ?  Ans.    iVV  ^  cu-  in. 


What  is  the  entire  area  of  the  surface  of  this  tetraedron  ? 


274 


ELEMENTARY    GEOMETRY, 


PROPOSITION    VII. 

632.  Theorem, — The  volume  of  the  frustum  of  a  tri- 
angular pyramid  is  equal  to  the  volume  of  three  -pyramids 
of  the  same  altitude  as  the  frustum,  and  whose  bases  are 
the  upper  base,  the  lower  base,  and  a  mean  proportional 
between  the  two  bases  of  the  frustum. 

Demonstration. 
Let  M&c-ABC  (Fig.  282)  be  the  frustum  of  a  triangular  pyramid. 

Through  db  and  C  pass  a  plane  cutting  off  the  pyramid  0-abc.  This 
has  for  its  base  the  upper  base  of  the  frustum,  and  for  its  altitude  the 
altitude  of  the  frustum. 

Again,  draw  Aft,  and  pass  a  plane  through 
Aft  and  ftC,  cutting  off  the  pyramid  ft-ABC, 
which  has  the  same  altitude  as  the  frustum, 
and  for  its  base  the  lower  base  of  the  frustum. 

There  now  remains  a  third  pyramid,  ft-AC«, 
to  be  examined. 

Through  ft  draw  ftD  parallel  to  «A,  and 
draw  DC  and  «D. 

The    pyramid    D-ACa    is    equivalent    to 
ft-ACrt,  since  it  has  the  same  base  and  the  same 
altitude  (?).     But  the  former  may  be  considered 
as  having  ADC  for  its  base,  and  the  altitude  of  the  frustum  for  its  altitude, 
i.  e.,  as  pyramid  «-ADC. 

We  are  now  to  show  that  ADC  is  a  mean  proportional  between  ahc 
and  ABC. 


Fig.  282, 


ABC 

abc 


AEr 


^1 

ad" 


(?). 


Also,  J— — 


ABC 
ADC 


AB 
AD 


(?);     or 


abc;;  ^  AT 

ADC'       ad' 


(?). 


Tj  ,.,      „     ,.        ABC       ABC 

By  equality  of  ratios,  — j—  = 5 ; 

'-''-        ADC' 


abc 


whence,     ADC"  =  aftcxABC 


i.e.^  ADC  is  a  mean  proportional  between  the  upper  and  lower  bases  of 
the  frustum. 

Hence  the  volume  of  the  frustum  is  equal  to  the  volume  of  three 
pyramids,  etc.     q.  e.  d. 


PYRAMIDS    AND    CONES.  275 

633.  Corollary. — The  volume  of  the  frustum  of  any 
pyramid  is  equal  to  the  volume  of  three  pyramids  hav- 
ing the  same  altitude  as  the  frustum,  and  for  bases,  the 
upper  base,  the  lower  base,  and  a  mean  proportional 
between  the  two  bases  of  the  frustum. 

For,  the  frustum  of  any  pyramid  is  equivalent  to  the  corresponding 
frustum  of  a  triangular  pyramid  of  the  same  altitude  and  an  equivalent 
base  (?) ;  and  the  bases  of  the  frustum  of  the  triangular  pyramid  being 
both  equivalent  to  the  corresponding  bases  of  the  given  frustum,  a  mean 
proportional  between  the  triangular  bases  is  a  mean  proportional  between 
their  equivalents. 


PROPOSITION    VIII. 

634.  Theorem. — The  volum^e  of  a  cone  of  revolution  is 
equal  to  one-third  the  product  of  its  base  and  altitude ; 
i.  e. ,  ^-n  R^H,  R  being  the  radius  of  the  base  and  H  the  alti- 
tude. 

Demonstration. 

The  volume  of  a  pyramid  is  equal  to  one-third  the  product  of  the 
base  and  altitude,  and  the  cone  being  the  limit  of  the  pyramid,  the  vol- 
ume of  the  cone  is  one-third  the  product  of  its  base  and  altitude. 

Now,  R  being  the  radius  of  the  base  of  a  cone  of  revolution,  the 
base  (area  of)  is  tt^,  whence  ^Tr^-fiTis  the  volume,  -S" being  the  altitude. 

Q.  E.  D. 

636.  Corollary  1. — The  volum^e  of  any  cone  is  equal 
to  one-third  the  product  of  its  base  and  altitude. 

636.  Corollary  2. — The  volume  of  the  frustum  of  a 
cone  is  equal  to  the  volum^e  of  three  cones  having  the 
same  altitude  as  the  frustum,,  and  for  bases,  the  upper 
base,  the  lower  base,  and  a  mean  proportional  between 
the  two  bases  of  the  frustum. 

The  truth  of  this  appears  from  the  fact  that  the  frustum  of  a  cone  is 
the  limit  of  the  frustum  of  a  regular  inscribed  pyramid. 


276  ELEMEJSTARY    GEOMETRY. 


PROPOSITION    IX. 

637.  Theorem. — The  lateral  surfaces  of  similar  pyra- 
mids are  to  each  other  as  the  squares  of  their  homologous 
edges,  or  of  their  altitudes. 

Demokstration. 

Let  A,  A',  A",  etc.,  and  a,  a\  a",  etc.,  be  homologous  sides  of  the 
bases  of  two  similar  pyramids,  E,  E',  E",  etc.,  and  e,  e',  e",  etc., 
homologous  lateral  edges,  H  and  h  the  altitudes  of  the  pyramids,  and 
let  S  and  &*  be  the  lateral  surfaces. 

C  A2  A/2  Kll-l 

_    ^    _    ^'    _    ^'        f 

~  e"   ~  e"    ~  e'""  '        ' 

Since  the  pyramids  are  similar,  the  corresponding  facial  angles  are 
equal,  and  the  homologous  edges  proportional  (597,  532),  hence  the 
bases  are  similar  polygons,  and  the  corresponding  lateral  faces  are  simi- 
lar triangles. 

Now  let  F,  F',  F",  etc.,  and/,/',/",  etc.,  be  the  corresponding  lateral 
faces,  of  which  triangles.  A,  A',  A",  etc.,  and  «,  a',  a'\  etc.,  are  the  bases 
respectively,  and  E,  E',  E",  etc.,  and  e,  e',  e",  etc.,  other  homologous  sides. 

A         A'         A"  F         F'         F"  U 

™™  *  =  *  =  *"  •  ^'<=-  =  J  =  7  =  ?-'  '^*<=-'  =  "  (')■ 

A*       A'"^        A"^  E^        E"        E'"  H^ 

^'^'^"'^^  a'=^  =  ^'  "^^  =  ¥  =  ^  =  7^'  «'<=•'  =  F  (^> 

F       A^      F'        A'2      F"        A"' 
Moreover, -^=^,    ^  =  -     _  =  _,  ,  etc. 

Whence  ^  +  ^'  +  ^"^  ^'^-  _  «  _  A^  _  A'«  _  A'"^ 

"^  i^  "^  ^  "^  e^ '  ®*^-'  ""  hF'  ^^^'^' 

638.  Corollary. — The  lateral  surfaces  of  similar  right 
pyramids  are  to  each  other  as  the  squares  of  any  homolo- 
gous lines,  as  slant  heights,  altitudes,  or  of  correspond- 
ing diagonals  of  the  bases.. 


PYRAMIDS    AND    CONES.  277 


PROPOSITION    X. 

639.  Theorem.  —  The  convex  surfaces  of  similar 
cones  of  revolution  are  to  each  other  as  the  squares  of 
their  slant  heights,  the  radii  of  their  bases,  or  their 
altitudes;  i.  e.,  as  the  squares  of  any  two  homologous 
dimensions. 

Demonstration. 

Let  H'  and  h'  be  the  slant  heights  of  two  similar  cones  of  revolution, 
R  and  r  the  radii  of  their  bases,  and  H  and  h  their  altitudes. 

Their  convex  surfaces  are  ttRH'  and  TrrA'. 


Now,  since  the  cones  are  similar, 


r        h>  ^'  '' 


Multiplying  the  terms  of  this  proportion    by  the    corresponding 
terms  of 

vH'  _  E' 

ttA'    ~  h'  ' 

.  irRH'       H'^ 

we  have  — ^7-  =  -^7^  • 

Hence  the  convex  surfaces  are  as  the  squares  of  their  slant  heights. 

Q.  E.  D. 


But,  as  _=_(?)  =  _, 


nRH^  _^  ^H* 

Trrh!   ~  r^  ~  h* 


That  is,  the  convex  surfaces  are  to  each  other  as  the  squares  of  the 
radii  of  the  bases,  or  as  the  squares  of  the  altitudes,    q.  e.  d. 


SJ78  ELEMENTARY    GEOMETRY, 


PROPOSITION    XI. 

640.  Theorem, — The  volumes  of  similar  pyramids 
are  to  each  other  as  the  cubes  of  their  homologous  dimen- 
sions. 

Synopsis  of  Demonstration. 

Let  A  and  a  be  homologous  sides  of  the  bases  of  two  similar  pyra- 
mids, B  and  b  their  bases,  and  J^and  h  their  altitudes. 

We  have 


B 

b 

=  f«' 

W 

P 

_A 
a 

=  > 

\BxH 

A^ 
~  a? 

=  f«- 

Q. 

B. 

D. 

PROPOSITION    XII. 

641.  Theorem. — The  volumes  of  similar  cones  are  to 
each  other  as  the  cubes  of  their  altitudes,  or  as  the  cubes 
of  the  radii  of  their  bases. 

Synopsis  of  Demonstration. 
Let  J?  and  r  be  the  radii  of  their  bases,  and  ^Tand  h  their  altitudes. 

We  have  ^  ^  'M^'^' 

••    ^Trr^xh    ~  h^^'' 

B? 
or  =^-    QE.D. 


REGULAR    POLYHDRONS.  279 


OF    THE    REGULAR    POLYEDRONS. 

642.  A  Polyedron  is  a  solid  bounded  by  plane  surfaces. 
A  Regular  Convex  Polyedron  is  a  polyedron  whose  faces  are  all 
equal  regular  polygons,  and  each  of  whose  solid  angles  is  convex 
outward,  and  is  enclosed  by  the  same  number  effaces. 


PROPOSITION    XIII. 

643.  Theorem.— There  are  five  and  only  five  regular 

convex  polyedrons,  viz. : 

The  TsTRAEDRONf  whose  faces  are  four  equal  equilat- 
eral  triangles ; 

The  Hexaedrox,  or  Cube,  whose  faces  are  six  equal 
squares ; 

The  OcTAEDRON,  whose  faces  are  eight  equal  equilateval 
triangles ; 

The  DoDECAEDRON,  whose  faces  are  twelve  equal  regular 
pentagons;  and 

The  IcosAEDRON,  whose  faces  are  twenty  equal  equilat- 
eral  triangles. 

Demonstration. 

We  demonstrate"  this  proposition  by  showing — Ist,  that  such  solids 
can  be  constructed ;  and  2d,  that  no  others  are  possible. 

The  Regular  Tetraedron.— Taking  three  equilat- 
eral triangles,  as  ASB,  ASC,  and  BSC,  it  is  possible  to 
enclose  a  solid  angle,  as  S,  with  them,  since  the  sum 
of  the  three  facial  angles  is  (what  ?)  (555). 

Then,  since  AC  =  AS  =  CB  (?),  considering  ACB 
the  fourth  face,  we  have  a  regular  polyedron  whose 
four  faces  are  equilateral  triangles.  ^'9-  ^^^- 


N 


1^80 


ELEMENTARY    GEOMETRY, 


The  Regular  Hexaedron  op  Cube.— This  is  a  familiar  solid,  but  for 
purposes  of  uniformity  and  completeness  we  may  conceive  it  constructefi 
as  follows:  Taking  three  equal  squares,  as  ASCB, 
CSED,  and  ASEF,  we  can  enclose  a  solid  angle,  as  S, 
with  them  (?). 

Now,  conceive  the  planes  of  CB  and  CD,  AB  and 
AF,  EF  and  ED  produced.  The  plane  of  CB  and  CD 
being  parallel  to  ASEF  (?),  will  intersect  the  plane  of 
EF  and  ED  in  HD  parallel  to  FE  (?).  In  like  manner, 
FH  can  be  shown  parallel  to  ED,  BH  to  CD,  and  HD  to 
BC.     Hence  the  solid  has  for  its  faces  six  equal  squares.         ^^^'  ^^ 


The  Regular  Octaedron. — At  the  intersection,  p, 
of  the  diagonals  of  a  square,  ABCD,  erect  a  perpendic- 
ular SP  to  the  plane  of  the  square,  and  making  SP  = 
AP  (half  of  one  of  the  diagonals)  draw  SA,  SD,  SC, 
and  SB. 

Making  a  similar  construction  on  the  other  side  of 
the  plane  ABCD,  we  have  a  solid  having  for  faces  eight 
equal  equilateral  triangles  (?). 


Fig.  185. 


The  Regular  Dodecaedron: — Taking  twelve  equal  regular  pentagons, 
we  may  group  them  in  two  sets  of  six  each,  as  in  the  figure.  Thus, 
around  0  we  may  place  five,  forming  five  triedrals  at  the  vertices  of  0. 
These  triedrals  are  possible, 
since  the  sum  of  the  facial 
angles  enclosing  each  is  3f 
righ!  angles  (?) — i.e. ,  between 
0  and  4  right  angles  (555). 

In  like  manner,  the  other 
six  may  be  grouped  by 
placing  five  of  them  about 
0'. 

XT  '    '       ^x.  '''9.  286. 

Now,  conceiving  the  con- 
vexity of  the  group  0  in  front  and  the  concavity  o^  grovip  0',  we  may  place 
the  two  together  so  as  to  inclose  a  solid.  Thus,  placing  A  at  &,  the  three 
faces  5,  7,  1,  will  inclose  a  triedral,  since  the  diedral  included  by  5  and  1 
is  the  diedral  of  such  a  triedral.  Then  will  vertex  B  fall  at  c,  and  a  like 
triedral  will  be  formed  at  that  point,  and  so  of  all  the  other  vertices. 
Hence  we  have  a  polyedron  having  for  faces  twelve  equal  regular  penta- 
gons. 


REG  ULAR     P  OL  YEDR  ONS. 


281 


Fig.  287. 


The  Regular  Icosaedron.— Taking  twenty  equal  equilateral  triangles, 
they  can  be  grouped  in  two  sets,  as  in  the  figure,  in  a  manner  altogether 
similar  to  the  preceding  case. 
The  solid  angles  in  this  case  are 
included  by  five  facial  angles 
whose  sum  is  3^  right  angles 
(:•),  which  is  a  possible  case 
(555).  As  before,  conceiving 
the  convexity  of  group  0  in 
front,  and  the  concavity  of  0', 
we  can  place  them  together  by 
placing  A  at  «,  thus  enclosing 
a  solid  angle  with  five  faces,  whence  B  will  fall  at  J,  etc.  Thus  we  obtain 
a  solid  with  twenty  equal  equilateral  triangles  for  its  faces. 

That  there  can  be  no  other  regular  polyedrons  than  these  five  is  evi- 
dent, since  we  can  form  no  other  convex  solid  angles  by  means  of  regular 
polygons.  Thus,  with  equilateral  triangles  (the  simplest  polygon)  we 
have  formed  solid  angles  with  three  faces  (the  least  number  possible),  as 
in  the  tetraedron ;  with  four,  as  in  the  octaedron ;  and  with  five,  as  in  the 
icosaedron.  Six  such  facial  angles  cannot  enclose  a  solfd  angle,  since 
their  sum  is  four  right  angles  (?),  and  much  less  can  any  greater  number. 
Again,  with  squares  (the  next  most  simple  polygon)  we  have  formed 
solid  angles  with  three  faces,  as  in  the  hexaedron,  and  can  form  no  other, 
for  the  same  reason  as  above.  With  regular  i)entagons  we  can  enclose 
only  a  triedral,  as  in  the  dodecaedron,  for  a  like  reason.  With  regular 
hexagons  we  cannot  enclose  a  solid  angle  (?),  and  much  less  with  any 
regular  polygon  of  more  than  six  sides. 


Fig.  288. 


282  ELEMENTARY    GEOMETRY. 

644.  Scholium. — Models  of  the  regular  polyedrons  are  easily  formed 
by  cutting  the  preceding  figures  from  cardboard,  cutting  half-way  through 
the  board  in  the  dotted  lines,  and  bringing  the  edges  together  as  the 
forms  will  readily  suggest. 


PROPOSITION    XIV. 

645.  Theorem. — Any  regular  polyedron  is  inscriptihle 
and  circumscriptible  by  a  sphere. 

Outline  of  Demonstration. 

From  the  centres  of  any  two  adjacent  faces,  as  c  and 
c',  let  fall  perpendiculars  upon  the  common  edge,  and 
they  will  meet  it  in  the  same  point  o  (?).  The  plane  of 
these  lines  will  be  perpendicular  to  this  edge  (?),  and 
perpendiculars  to  these  faces  from  their  centres,  as  cS, 
c'S,  will  lie  in  this  plane  (?),  and  hence  will  intersect  at 
a  point  equally  distant  from  these  faces  (?). 

In  like  manner  c"S  =  c'S,  and  the  point  S  can  be  ^'^"  ^^^' 

shown  to  be  equally  distant  from  all  of  the  faces,  and  is  therefore  the  cen- 
tre of  the  inscribed  sphere. 

Joining  S  with  the  vertices,  we  can  readily  show  that  S  is  also  the 
centre  of  the  circumscribed  sphere. 


EXERCISES. 

646.  1.  What  is  the  area  of  the  lateral  surface  of  a  right 
hexagonal  pyramid  whose  base  is  inscribed  in  a  circle  whose 
diameter  is  20  feet,  the  altitude  of  the  pyramid  being  8  feet  ? 
What  is  the  volume  of  this  pyramid  ? 

2.  What  is  the  area  of  the  lateral  surface  of  a  right  pentago- 
nal pyramid  whose  base  is  inscribed  in  a  circle  whose  radius  is 
G  yards,  the  slant  height  of  the  pyramid  being  10  yards?  What 
is  the  volume  of  this  pyramid  ? 


EXERCISES.  383 

3.  How  many  quarts  will  a  can  contain,  whose  entire  height 
is  10  inches,  the  body  being  a  cylinder  6  inches  in  diameter  and 
Q^  inches  high,  and  the  top  a  cone?  How  much  tin  does  it  take 
to  make  such  a  can,  allowing  nothing  for  waste  and  the  seams  ? 

4.  If  very  fine  dry  sand  is  piled  upon  a  smooth  horizontal 
surface,  without  any  lateral  support,  the  angle  of  slope  (i.  e.,  the 
angle  of  inclination  of  the  sloping  side  of  the  pile  with  the  plane) 
is  about  31°.  Suppose  two  circles  be  drawn  on  the  floor,  one 
4  feet  in  diameter  and  the  other  3,  and  sand  piles  be  made  as 
large  as  possible  on  these  circles  as  bases,  no  other  support  being 
giyen.     What  is  the  relative  magnitude  of  the  piles  ? 

5.  In  the  case  of  sand  piles,  as  given  in  the  last  example,  the 
ratio  of  the  radius  of  the  base  to  the  altitude  of  the  pile  is  |. 
How  many  cubic  feet  in  each  of  the  above  piles  ? 

6.  The  frustum  of  a  right  pyramid  was  72  feet  square  at  the 
lower  base  and  48  at  the  upper ;  and  its  altitude  was  60  feet. 
What  was  the  lateral  surface  ?  What  the  volume  ?  [Such  a 
solid  is  called  a  Prismoid.] 

7.  Find  the  area  of  the  surface,  and  the  contents  of  a  regular 
tetraedron,  one  of  whose  edges  is  10  inches.  What  is  the  diam- 
eter of  the  inscribed  sphere  ?    Of  the  circumscribed  ? 

647.  A  Wedge  is  a  sohd 
bounded  by  three  quadrilaterals 
and  two  triangles. 

Thus,  ABCD  is  a  rectangle,  and  is 
called  the  Head  of  the  wedge,  the  two 
Irianj^les  AED  and  FBC  are  the  Ends, 
and  the  two  trapezoids  ABFE  and 
DCFE  are  the  Sides.  The  Altitude  is 
the  perpendicular  to  the  head  from  the  Fiq.  290. 

edge  opposite. 

8.  The  base  of  a  wedge  being  18  feet  by  9  feet,  the  edge  20 
feet,  and  the  altitude  6  feet,  what  are  the  contents  ? 

Ans.  504  cu.  ft. 


284 


ELEMENTARY    GEOMETRY. 


OF    THE    SPHERE.* 

648.  A  Sphere  is  a  solid  bounded  by  a  surface  every  point 
in  which  is  equally  distant  from  a  point  within  called  the  Cejitre. 

The  distance  from  the  centre  to  the  surface  is  the  Radius, 
and  a  line  passing  through  the  centre  and  limited  by  the  surface 
is  a  Diameter.     The  diameter  is  equal  to  twice  the  radius. 


CIRCLES    OF    THE    SPHERE, 


PROPOSITION    I. 

649.    Theorem. — Every  section  of  a  sphere  made  hy  a 
plane  is  a  circle. 

Demonstration. 

Let  AFEBD  be  a  section  of  a  sphere, 
whose  centre  is  0,  made  by  a  plane;  then 
is  the  section  AFEBD  a  circle. 

For,  let  fall  from  the  centre  0  a  per- 
pendicular upon  the  plane  AFEBD,  as  OC, 
and  draw  CA,  CD,  CE,  CB,  etc.,  lines  of  the 
plane,  from  the  foot  of  the  perpendicular 
to  any  points  in  which  the  plane  cuts  the  Fig.  29i. 


*  A  spherical  blackboard  is  almost  indispensable  in  teaching  this  section 
as  well  as  in  teaching  Spherical  Trigonometry.  A  sphere  about  two  feet  in 
diameter,  mounted  on  a  pedestal,  and  having  its  surface  slated  or  painted  as 
a  blackboard,  is  what  is  needed.  It  can  be  obtained  of  the  manufacturers 
of  school  apparatus,  or  made  in  any  good  turning-shop. 


OF    THE    SPHERE.  285 

surface  of  the  sphere.     Join   these  points  with   the   centre,  0,  of  the 
sphere. 

Now  OA,  OD,  OB,  OE,  etc.,  being  radii,  are  equal;  whence,  CA,  CD, 
CB,  CE,  etc.,  are  equal;  i.  e.,  every  point  in  the  line  of  intersection  of  a 
plane  and  surface  of  a  sphere  is  equally  distant  from  a  point  in  this 
plane.     Hence,  the  intersection  is  a  circle,     q.  e.  d. 

650.  A  circle  made  by  a  plane  not  passing  through  the  centre 
is  a  Small  Circle ;  one  made  by  a  plane  passing  through  the 
centre  is  a  Great  Circle. 

651.  Corollary  \.—A  perpendicidar  from  the  centre 
of  a  sphere  upon  any  small  circle  pierces  the  cii'de  at  its 
centre;  and,  conversely,  a  perpendicular  to  a  small  circle 
at  its  centre  passes  through  the  centre  of  the  sphere. 

652.  A  diameter  perpendicular  to  any  circle  of  a  sphere  is 
called  the  Axis  of  that  circle.  The  extremities  of  the  axis  are 
the  Poles  of  the  circle. 

663.   Corollary  2. — The  pole  of  a  circle  is  equally  dis- 
tant from  every  point  in  its  circumference. 
The  student  should  give  the  reason. 

654.  Corollary  3. — Every  circle  of  a  sphere  has  two 
poles,  which,  in  case  of  a  great  circle,  are  equally  distant 
from  every  point  in  the  circumference  of  the  circle ;  hut, 
in  case  of  a  small  circle,  one  pole  is  nearer  any  point  in 
the  circumference  than  the  other  pole  is. 

655.  Corollary  4. — A  small  circle  is  less  a;  it.<  dis- 
tance from  the  centre  of  the  sphere  is  greater ;  )  enc£  tlie 
circle  whose  plane  passes  through  the  centre  is  th  greatest 
circle  of  the  sphere. 

For,  its  diameter,  being  a  chord  of  a  great  circle,  is  less  as  it  i 
ther  from  the  centre  of  the  great  circle,  which  is  also  the  cenfrr     i  <   . 
sphere. 

656.  Corollary  5. — All  great  circles  of  the  same  sphere 
are  equal  (?). 


286  ELEMENTARY     GEOMETRY. 


PROPOSITION    II. 

667.  Theorem. — Any  great  circle  divides  the  sphere 
into  two  equal  parts. 

DEMOlfSTRATIOlir. 

Conceive  a  sphere  as  divided  by  a  great  circle,  i.  e.,  by  a  plane  pass- 
ing through  its  centre,  and  let  the  great  circle  be  considered  as  the  base 
of  each  portion.  These  bases  being  equal,  reverse  one  of  the  portions  and 
conceive  its  base  placed  in  the  base  of  the  other,  the  convex  surfaces 
being  on  the  same  side  of  the  common  base.  Since  the  bases  are  equal 
circles,  they  will  coincide,  and  since  all  points  in  the  convex  surface  of 
each  portion  are  equally  distant  from  the  centre  of  the  common  base,  the 
convex  surfaces  will  coincide.  Therefore,  the  portions  coincide  through- 
out, and  are  consequently  equal.     Q.  e.  d. 

657,  a.— A  Hemisphere  is  one  of  the  two  equal  parts  into 
which  a  great  circle  divides  a  sphere. 


PROPOSITION    III. 

658.  Theorem.— The  intersection  of  any  two  great  cir- 
cles of  a  sphere  is  a  diameter  of  a  sphere. 

Demonstration. 

The  intersection  of  two  planes  is  a  straight  line ;  and  in  the  case  of 
the  two  great  circles,  as  they  both  pass  through  the  centre  of  the  sphere, 
this  is  one  point  of  their  intersection.  Hence,  the  intersection  of  two 
great  circles  of  a  sphere  is  a  straight  line  which  passes  through  the  cen- 
tre.    Q.  E.  D. 

659.  Corollary. — The  intersections  on  the  surface  of  a 
sphere  of  two  circumferences  of  great  circles  are  a  semi 
circumference,  or  180°,  apart,  since  they  are  '^*  /..r,v.c// 
extremities  of  a  diameter. 


OF    THE    SPHERE. 


287 


DISTANCES    ON    THE    SURFACE    OF    A 
SPHERE. 

660.  Distances  on  the  surface  of  a  sphere  are  always  to  be 
understood  as  measured  on  the  arc  of  a  great  circle,  unless  it  is 
otherwise  stated. 


PROPOSITION    IV. 

661.  Theorem. — The  distance,  measured  on  the  sur- 
face of  a  sphere,  from  the  pole  of  a  circle  to  any  point  in 
the  circumference  of  that  circle,  is  the  same. 

Demonstration. 

Let  P  be  a  pole  of  the  small  circle  AEB. 

Then  are  the  arcs  PA,  PE,  PB,  etc., 
which  measure  the  distances  on  the  surface 
of  tlie  sphere,  from  P  to  any  points  in  the 
circumference  of  circle  AEB,  equal. 

For,  by  (653),  the  straight  lines  AP,  PE, 
PP,  etc.,  are  equal,  and  these  equal  chords 
subtend  equal  arcs,  as  arc  PA,^rc  PE,  arc 
BB,  etc.,  the  great  circles  of  which  these 
lines  are  chords  and  arcs  being  equal  (656). 

Thus,  for  like  reasons, 


Fig.  292. 


arc  P'QA  =  arc  P'LE  =  arc  P'RB,  etc.     q.  e.  d. 


622.  Corollary. — The  distance  from  the  pole  of  a  great 
circle  to  any  point  in  the  circumference  of  the  circle  is  a 
quadrant  (a  quarter  of  a  circumference). 

Since  the  poles  are  180°  apart  (being  the  extremities  of  a  diameter), 
PAQP'  =  PELP'  =  a  serai-circumference.  But,  in  case  of  a  great  circle, 
chord  PL  =  chord  P'L  (=  chord  PQ  =  chord  P'Q),  whence  arc  PEL  = 
arc  P'L  =  arc  PAQ  =  arc  P'Q.    Hence,  each  of  these  arcs  is  a  quadrant. 


288 


•  ELEMENTARY    GEOMETRY. 


663.  Scholium. — By  means  of  the  facts 
demonstrated  in  this  proposition  and  corollary, 
we  are  enabled  to  draw  arcs  of  small  and  great 
circles,  in  the  surface  of  a  sphere,  with  nearly  the 
same  facility  that  we  draw  arcs  and  lines  in  a 
plane.  Thus,  to  draw  the  small  circle  AEB  (Fig. 
292),  we  take  an  arc  equal  to  PE,  and  placing  one 
end  of  it  at  P,  cause  a  pencil  held  at  the  other 
end  to  trace  the  arc  AEB,  etc.  To  describe  the 
circumference  of  a  great  circle,  a  quadrant  must 
be  used  for  the  arc.  By  bending  a  wire  into  an  arc  of  the  circle,  and 
making  a  loop  in  each  end,  a  wooden  pin  can  be  put  through  one  loop 
and  a  crayon  through  the  other,  and  an  arc  drawn  as  represented  in 
Fig.  293. 


Fig.  293. 


PROPOSITION     V. 

664.    Problem. — To  pass  a  circumference  of  a  great 
circle  through  any  two  points  on  the  surface  of  a  sphere. 


Solution. 

Let  A  and  B  |3e  two  points  on  the  surface  of  a  sphere,  through  which 
it  is  proposed  to  pass  a  circumference  o/a  great  circle. 

From  B  as  a  pole,  with  an  arc  equal  to  a  quad- 
rant, strike  an  arc  on,  as  nearly  where  the  pole  of 
the  circle  passing  through  A  and  B  lies,  as  may  be 
determined  by  inspection.  Then,  from  A,  with 
the  same  arc,  strike  an  arc  st  intersecting  on  at  P. 
Now,  P  is  the  pole  of  the  great  circle  passing 
through  A  and  B  (?).  Hence,  from  P  as  a  pole,  with 
a  quadrant  arc  drawing  a  circle,  it  will  pass 
through  A  and  B ;  and  it  will  be  a  great  circle, 
since  its  pole  is  a  quadrant's  distance  from  its  circumference. 

[The  student  should  make  this  construction  on  the  spherical  black- 
board.] 


Fig.  294. 


OF    THE    SPHERE.  289 


PROPOSITION    VI. 

665.  Theorem, — Through  any  two  points  on  the  sur- 
face of  a  sphere,  one  great  circle^  can  always  he  made  to 
pass,  and  only  one,  except  when  the  two  points  are  at  the 
extremities  of  the  sam^e  diameter,  in  which  case  an  infi- 
nite number  of  great  circles  can  he  passed  through  the  two 
points. 

Demonstration. 

This  proposition  may  be  considered  a  corollary 
to  the  preceding.  Thus,  in  general,  the  two  great 
circles  struck  from  A  and  B  as  poles,  with  a  quad- 
rant arc,  can  intersect  in  only  two  points  (?),  which 
are  the  poles  of  the  same  great  circle  (?). 

But,  if  the  two  given  points  were  at  the  extrem- 
ities of  the  same  diameter,  as  at  D  and  C,  the  arcs 
st  and  m  would  coincide,  and  any  point  in  this  ^'8-  295. 

circumference  being  takdn  as  a  pole,  great  circles  can  be  drawn  through 
D  and  C. 

[The  student  should  trace  the  work  on  the  spherical  blackboard.] 

666.  Scholium. — The  truth  of  the  proposition  is  also  evident  from 
the  fact  that  three  points  not  in  the  same  straight  line  determine  the 
position  of  a  plane.  Thus.  A,  B,  and  the  centre  of  the  sphere,  fix  the 
position  of  one,  and  only  one,  great  circle  passing  through  A  and  B. 
Moreover,  if  the  two  given  points  are  at  the  extremities  of  the  same  diam- 
eter, they  are  in  the  same  straight  line  with  the  centre  of  the  sphere, 
whence  an  infinite  number  of  planes  can  be  passed  through  them  and  the 
centre.  The  meridians  on  the  earth's  surface  afford  an  example,  the  poles 
(of  the  equator)  being  the  given  points. 

667.  Corollary. — //  two  points  in  the  circumference  of 
a  great  circle  of  a  sphere,  not  at  the  extremities  of  the 
sam^e  diameter,  are  at  a  quadrant's  distance  from  a  point 
on  the  surface,  this  point  is  the  pole  of  the  circle, 

"■  The  word  circle  may  be  understood  to  refer  either  to  the  circle  proper, 
or  to  its  circumference.     The  word  is  in  constant  use  in  the  higher  mathe- 
matics in  the  latter  sense. 
13 


290  ELEMENTARY    GEOMETRY. 


PROPOSITION    VII. 

668.  Theorem. — The  shortest  distance  on  the  surface 
of  a  sphere,  between  any  two  points  in  that  surface,  is 
measured  on  the  arc  less  than  a  semi-circumference  of  the 
great  circle  which  joins  them. 

Demonstration. 

Let  A  and  B  be  two  points  in  the  surface  of  a  sphere,  AB  the  arc  of  a 
great  circle  joining  them,  and  kiuCn^  any  other  path  in  the  surface  be- 
tween A  and  B. 

Then  is  arc  AB  less  than  AmCwB. 

Let  C  be  any  point  in  AmCwB,  and  pass  the 
arcs  of  great  circles  through  A  and  C,  and  B  and 
C.  Join  A,  B,  and  C  with  the  centre  of  the  sphere. 
The  angles  AOB,  AOC,  and  COB  form  the  facial 
angles  of  a  triedral,  of  which  angles  the  arcs  AB, 
AC,  and  CB  are  the  measures. 

Now,    angle  AOB  <  AOC  +  COB  (540) ;  Fig.  296 

whence  arc  AB  <  arc  AC  +  arc  CB  (?), 

and  the  path  from  A  to  B  is  less  on  arc  AB  than  on  arcs  AC,  CB. 

In  like  manner,  joining  any  point  in  AmC  with  A  and  C  by  arcs  of 
great  circles,  their  sum  will  be  greater  than  AC.  So,  also,  joining  any 
point  in  CwB  with  C  and  B,  the  sum  of  the  arcs  will  be  greater  than  CB. 

As  this  process  is  indefinitely  repeated,  the  path  from  A  to  B  on  the 
arcs  of  the  great  circles  will  continually  increase,  and  also  continually 
approximate  the  path  AmCwB.  Hence,  arc  AB  is  less  than  the  path 
AmCwB.     Q.  E.  D. 

669.  Corollary. — The  least  arc  of  a  circle  of  a  sphere 
joining  any  two  points  in  the  surface,  is  the  arc  less  than 
a  semi-circumference  of  the  great  circle  passing  through 
the  points;  and  the  greatest  arc  is  the  circumference 
minus  this  least  arc. 


OF   TEE    SPHERE. 


291 


Thus,  let  Km^n  be  any  small  circle  passing 
through  A  and  B,  and  ABD^^C  the  great  circle; 
then,  as  just  shown,  Ajt>B  <  A77iB. 

Now,   circf.*ABDoC  >  circf.  AmB/i  (655). 

Subtracting  the  former  inequality  from  the 
latter,  we  have  BDoCA  >  B/iA.    Q.  e.  d. 

670.  Two  arcs  of  great  circles  are  said 
to  be  perpendicular  to  each  other  when 
their  circles  are. 


Fig.  297. 


•  PROPOSITION    VIII. 

671.  Theorem. — If  at  the  middle  point  of  an  aro  of  a 
great  circle  a  perpendicular  is  drawn  on  the  surface  of  a 
sphere,  the  distances  being  measin^ed  on  great  circles, 

1st.  Any  p9^^  ^^  ^^^^  perpendicular  is  equally  distant 
from  the  extremities  of  the  arc, 

2d.  Anij  point  out  of  the  perpendicular  is  unequally 
distant  from  the  extremities  of  the  arc. 


Demonstration". 

Let  AB  be  any  arc  of  a  great  circle,  D  its  middle  point,  and  PD  a 
perpendicular. 

Then  is  PB  =  PA,  the  ares  being  all  arcs  of 
great  circles. 

From  0,  the  centre  of  the  sphere,  draw  OP, 
OD,  OB,  and  OA.  The  rectangular  triedrals 
0-PDB  and  0-PDA  are  symmetrically  equal  (?) ; 
whence  PB  =  PA.     Q.  e.  d. 


Fig.  298. 


Again,  let  P'  be  a  point  out  of  PD.  Pass 
arcs  of  great  circles  through  P'  and  A,  and  P'  and 
B,  as  P'A,  P'B.  From  P,  where  one  of  these 
cuts  PD,  draw  the  arc  of  a  great  circle  PB.     Then  is 

P'B  <  P'P  +  PB  (668), 
whence,  P'B  <  P'P  +  PA  (?),  and  P'B  <  P'A  (?).     Q.  e.  d. 


292 


ELEMENTARY    GEOMETRY. 


672.  Corollary  1. — The  perpendicular  at  the  middle 
point  of  an  arc  contains  all  the  points  in  the  surface  of 
the  sphere  which  are  equally  distant  from  tJJn>e  extremities 
of  the  arc, 

673.  Corollary  2. — An  arc  which  has  each  of  two  points, 
not  at  the  ejotremity  of  the  same  diameter,  equally  distant 
from  the  extremities  of  another  arc  of  a  great  circle,  is 
perpendicular  to  the  latter  at  its  middle  point. 

This  is  apparent,  since  by  Corollary  1  such  points  are  in  the  perpen- 
dicular, and  two  such  points  with  the  centre  determine  a  great  circle. 


PROPOSITION    IX. 

674.  Theorem. — The  shortest  path  on  the  surface  of  a 
hemisphere,  from  any  point  therein  to  the  circumference 
of  the  great  circle  forming  its  base,  is  the  arc  not  greater 
than  a  quadrant  of  a  great  circle  perpendicular  to  the 
base,  and  the  longest  path,  on  any  arc  of  a  great  circle,  is 
the  supplement  of  this  shortest  path* 


Dbmokstration. 

Let  P  be  a  point  in  the  surface  of  the  hemisphere  whose  base  is 
ADCBC,  and  DPmD'  an  arc  of  a  great  circle  passing  through  P  and  per- 
pendicular to  ADCBC. 

Then  is  PD  the  shortest  path  on  the  sur- 
face from  P  to  circumference  ADCBC,  and 
P?7iD'  is  the  longest  path  from  P  to  the  cir- 
cumference, measured  on  the  arc  of  a  great 
circle. 

For,  the  shortest  path  from  P  to  any  point 
in  circumference  ADBC  is  measured  on  the 
arc  of  a  great  circle  (?).  Now,  let  PC  be  any 
oblique  arc  of  a  great  circle.  We  will  show 
that  Fig.  29! 

arc  PD  <  arc  PC. 


OF    THE    SPHERE. 


293 


Produce  PD  until  DP'  =  PD;  and  pass  a  great  circle  through  P' 
and  C. 

Then  is  the  arc  PC  =  arc  P'C. 

And,  since  PC  +  PC  >  PP', 

PC,  the  half  of  PC  4-  P'C,  is  greater  than  PD,  the  half  of  PP'.  q.  e.  d. 

Secondly,  PmD'  is  the  supplement  of  PD,  and  we  are  to  show  that  it 
Ml  greater  than  any  other  arc  of  a  great  circle  from  P  to  the  circumference 
ADBC.  Let  FnC  be  any  arc  of  a  great  circle  oblique  to  ADCBC.  Pro- 
duce C'wP  to  C.  Now  CPtiC  is  a  semi-circumference  and  consequently 
equal  to  DPmD'.    But  we  have  before  shown  that 

PD  <  PC, 

and  subtracting  these  from  the  equals  CPnC  and  DPwD',  we  have 

PwD'  >  PnC.   Q.  E.  D. 

675.  Corollary. — From  any  point  in  the  surface  of  a 
hemisphere  there  are  two  perpendiculars  to  the  circumfer- 
ence of  the  great  circle  which  forms  the  base  of  the  hemi- 
sphere; one  of  which  perpendiculars  measures  the  least 
distance  to  that  circumference,  and  the  other  the  greatest, 
on  the  arc  of  any  great  circle  of  the  sphere. 


SPHERICAL    ANGLES. 


676.  The  angle  formed  by  two  arcs 
of  circles  of  a  sphere  is  conceived  as 
the  same  as  the  angle  included  by  the 
tangents  to  the  arcs  at  the  common 
points. 

Illubtbation. — Let  AB  and  AC  be  two 
arcs  of  circles  of  the  sphere,  meeting  at  A  ; 
then  the  angle  BAC  is  conceived  as  the  same 
as  the  angle  B'AC,  B'A  being  tangent  to  the 
circle  BADm,  and  C'A  to  the  circle  CAEn. 


Fio.  300. 


294 


ELEMENTARY     GEOMETRY. 


677.  A  Spherical  Angle  is  the  angle  included  by  two 
arcs  of  grmt  circles. 

Illustration.— B AC  is  a  spherical  an- 
gle, and  is  conceived  as  the  same  as  the 
angle  B'AC,  B'A  and  C'A  being  tangents  to 
the  great  circles  BADF  and  CAEF.  [The  stu- 
dent should  not  confound  such  an  angle  as 
BAC  Fig.  800)  with  a  spherical  angle.] 

Fig.  301. 


PROPOSITION    X. 

678.  Theorem. — td  spherical  angle  is  equal  to  the 
measicre  of  the  diedral  included  by  the  great  circles  whose 
arcs  form  the  sides  of  the  angle. 


Demon^stration. 

Let  BAC  be  any  spherical  angle,  and  BADF  and  CAEF  the  great  cir- 
cles whose  arcs  BA  and  CA  Include  the  angle. 


Then  is  BAC  equal  to  the  measure  of  the 
diedral  C-AF-B. 

For,  since  two  great  circles  interaect  in  a 
diameter  (?),  AF  is  a  diameter. 

Now  B'A  is  a  tangent  to  the  circle  BADF, 
that  is,  it  lies  in  the  same  plane  and  is  per- 
pendicular to  AO  at  A. 

In  the  like  manner,  C'A  lies  in  the  plane 
CAEF  and  is  perpendicular  to  AO.  Hence 
B'AC'  is  the  measure  of  the  diedral  C-AF-B  (?). 


Fig.  302. 


Therefore  the  spherical  angle  BAC,  which  is  the  same  as  the  plane 
angle  B'Ab',  is  equal  to  the  measure  of  the  diedral  C-AF-B.    q.  e.  d. 


OF    THE    UPHERE, 


295 


PROPOSITION    XI 

679.  Theorem.—//  one  of  two  great  circles  passes 
through  the  pole  of  the  other,  their  circinriferences  inter- 
sect at  right  angles. 

Demonstration. 


Thus,  P  being  the  pole  of  the  great  circle 
CABm,  PO  is  its  axis,  and  any  plane  passing 
through  PO  is  perpendicular  to  the  plane 
CABm.  (?). 

Hence,  the  diedral  B-AO-P  is  right,  and 
the  spherical  angle  PAB,  which  is  equal  to  the 
measure  of  the  diedral,  is  also  right,     q.  e.  d. 


Fig.  303. 


680.  Corollary  1. — A  spherical  angle  is  measured  by 
the  arc  of  a  great  circle  intercepted  between  its  sides,  and 
at  a  quadrant's  distance  from  its  vertex. 

Thus,  the  spherical  angle  CPA  is  measured  by  CA,  PC  and  PA  being 
quadrants.  For,  since  PC  is  a  quadrant,  CO  is  a  perpendicular  to  PO, 
the  edge  of  the  diedral  C-PO-A,  and  for  the  like  reason  AO  is  perpendic- 
ular to  PO.  Hence,  COA  is  the  measure  of  the  diedral,  and  consequently 
CA,  its  measure,  is  the  measure  of  the  spherical  angle  CPA.    q.  k.  d. 

681.  Corollary  2. — The  angle  included  by  two  arcs  of 
sm,all  circles  is  the  same  as  the  angle  included,  bi/  two  arcs 
of  great  circles  passing  through  the  vertex  and  having  the 
same  .tangents. 


Thus, 


BAC  =  B  AC 


For  the  angle  BAC  is,  by  definition,  the 
same  as  B'AC,  B'A  and  CA  being  tangents 
to  BA  and  CA.  Now,  passing  planes 
through  CA,  B'A.  and  the  centre  of  the 
sphere,  we  have  the  arcs  B"A,  C'A,  and  B'A, 
C'A  tangents  to  them.  Hence,  B"AC"  is 
the  same  as  B'AC,  and  consequently  the 
same  as  BAC.    q.  e.  d. 


Fig.    304. 


296 


ELEMENTARY    GEOMETRY, 


682.  Scholium. — To  draw  an  arc  of  a  great  circle  which 
shall  he  perpendicular  to  another ;  or,  what  is  the  same 
thing,  to  construct  a  right  spherical  angle. 

Let  it  be  required  to  erect  an  arc  of  a  great  circle  perpendicular  to 
CAB  at  A.  Lay  oflFfrom  A,  on  the  arc  CAB, 
a  quadrant's  distance,  as  AP',  and  from  P'  as 
a  pole,  with  a  quadrant  describe  an  arc  pass- 
ing through  A.  This  will  be  the  perpendic- 
ular required. 

In  a  similar  manner  we  may  let  fall  a  per- 
pendicular from  any  point  in  the  surface, 
upon  any  arc  of  a  great  circle.  To  let  fall  a 
pei-pendicular  from  P"  upon  the  arc  CAB, 
from  P"  as  a  pole,  with  a  quadrant  describe 
an  arc  cutting  CAB,  as  at  P'.     Then,  from  P' 

as  a  pole,  with  a  quadrant  describe  an  arc  passing  through  P"  and  cutting 
CAB,  and  it  will  be  perpendicular  to  CAB. 


Fig.  305 


PROPOSITION    XII. 

683.  Problem. — To  pass  the  circumference  of  a  small 
circle  through  any  three  points  on  the  surface  of  a  sphere. 

Solution-. 

Let  A,  B,  and  C  be  the  three  points  in  the  surface  of  the  sphere 
through  which  we  propose  to  pass  the  circumference  of  a  circle. 

Pass  arcs  of  great  circles  through  the  points, 
thus  forming  the  spherical  triangle  ABC  (664)- 

Bisect  two  of  these  arcs,  as  BC  and  AC,  by 
arcs  of  great  circles  perpendicular  to  each  (673, 
682)-  The  intersection  of  these  perpendiculars, 
t>,  will  be  the  pole  of  the  small  circle  required  (?). 

Then  from  <?,  as  a  pole,  with  an  arc  oB  draw 
the  circumference  of  a  small  circle:  it  will  pass 
through  A,  B,  and  C  (?),  and  hence  is  the  circum- 
ference required. 

QpERY.  —If  the  three  given  points  chance  to  be  in  the  circumference 
of  a  great  circle,  how  will  it  appear  in  the  construction? 


Fig.  306. 


OF    THE    SPHERE, 


297 


OF    TANGENT    PLANES. 


684.  A  Tangent  Plane  to  a  curved  surface  at  a  given 
point  is  the  plane  of  two  lines  respectively  tangent  to  two  plane 
sections  through  the  point. 

Illustration. — Let  P  be  any 
point  in  the  curved  surface.  Pass  any 
two  planes  through  the  surface  and 
the  point  P,  and  let  OPQ  and  MPN 
represent  the  intersections  of  these 
planes  with  the  curved  surface. 
Draw  UV  and  ST  in  the  planes  of 

the  sections,  and  tangent  respectively  to  OPQ  and  MPN  at  P. 
the  plane  of  UV  aud  ST  the  tangent  plane  at  P. 


Fig.  307. 


Then  is 


PROPOSITION    XIII. 

686.    Theorem. — «^  tangent  plane  to  a  sphere  is  per- 
pendicular  to  the  radiios  at  the  point  of  tangency. 


Demonstration. 

Let  P  be  any  point  in  the  surface  of  a  sphere ;  pass  two  great  circles, 
as  Pa  A,  etc.,  and  P///AR,  through  P,  and  draw  ST  tangent  to  the  arc 
mP,  and  UV  tangent  to  the  arc  a?. 

Then  is  the  plane  SVTU  a  tangent 
plane  at  P,  and  perpendicular  to  the  ra- 
dius OP. 

For,  a  tangent  (as  ST)  to  the  arc  mP 
is  perpendicular  to  the  radius  of  the  cir- 
cle, i.  e.,  to  OP,  and  also  a  tangent  (as  VU) 
to  the  arc  AP  is  perpendicular  to  the  ra- 
dius of  tliis  circle,  i.  e.,  to  OP. 

Hence,  OP  is  perpendicular  to  two 
lines  of  the  plane  SVTU,  and  consequent- 
ly to  the  plane  of  these  lines  (?).    q.  e.  d. 


ELEMENTARY    GEOMETRY. 

686.  OoBOLLARY  1. — Jijvery  point  in  a  tangent  plane  to 
a  sphere,  except  the  point  of  tangency,  is  without  the 
spherB, 

For,  OP,  the  perpendicular,  is  shorter  than  any  line  which  can  be 
drawn  from  0  to  any  other  point  in  the  plane  (?)  ;  hence  any  other  point 
in  the  plane  than  P  lies  farther  from  the  centre  of  the  sphere  than  the 
length  of  the  radius,  and  is,  therefore,  without  the  sphere. 

687.  Corollary  2. — A  tangent  through  P  to  any  circle 
of  the  sphere  passing  through  this  point  lies  in  the  tan- 
gent plane* 

Thus,  MN,  tangent  to  the  small  circle  PwRft  through  P,  lies  in  the  tan- 
gent plane. 

For,  conceive  the  plane  of  the  small  circle  extended  till  it  intersects 
the  tangent  plane.  This  intersection  is  tangent  to  the  small  circle,  since 
it  touches  at  one  point,  but  cannot  cut  it ;  otherwise  the  tangent  plane 
would  have  another  point  than  P  common  with  the  surface  of  the  sphere. 

But  there  can  be  only  one  tangent  to  a  circle  at  a  given  point.  Hence 
this  intersection  is  MN,  which  is  consequently  in  the  tangent  plane. 


OF    SPHERICAL    TRIANGLES. 

688.  A  Spherical  Triangle  is  a  portion  of  the  surface  of 
a  sphere  bounded  by  three  arcs  of  great  circles.  In  the  present 
treatise  these  arcs  will  be  considered  as  each  less  than  a  semi- 
circumference;  and  the  triangle  considered  will  be  the  one  which 
is  less  than  a  hemisphere. 

The  terms  scalene,  isosceles,  equilateral,  right-angled,  and 
oblique-angled,  are  applied  to  spherical  triangles  in  the  same 
manner  as  to  plane  triangles. 


OF    THE    SPHERE. 


299 


PROPOSITION    XIV. 

Theorem. — The  sum  of  any  two  sides  of  a 
spherical  triangle  is  greater  than  the  third  side,  and 
their  difference  is  less  than  the  third  side. 


Demonstration. 
Let  ABC  be  any  spherical  triangle. 
Then  is  BC  <  BA  +  AC, 

and  BC  —  AC  <  BA ; 

and  the  same  is  true  of  the  sides  in  any  order. 

For,  join  the  vertices  A,  B,  and  C  with  the  cen- 
tre of  the  sphere,  by  drawing  AO,  BO,  and  CO. 
There  is  thus  formed  a  triedral    0-ABC,  whose  '"'S-  ^°^- 

facial  angles  are  measured  by  the  sides  of  the  triangle  (188).  Now, 
angle  BOC  is  less  than  BOA  +  AOC  (?),  whence  BC  is  less  than  BA  +  AC ; 
and  substracting  AC  from  each  member,  we  have  BC— AC  less  than  BA. 

Q.  E.  D. 


PROPOSITION     XV. 

690.  Theorem. — The  sum  of  the  sides  of  a  spherical 
triangle  may  be  anything  between  0  and  a  circumfer- 
ence' 

Demonstration. 

The  sides  of  a  spherical  triangle  are  measures  of  the  facial  angles  of  a 
triedral  whose  vertex  is  at  the  centre  of  the  sphere.  Hence  their  sum 
may  be  anything  between  0  and  the  measure  of  4  right  angles,  as  these 
are  the  limits  of  the  sum  of  the  facial  angles  of  a  triedral  (?).  q.  e.  d. 

691.  Scholium. — As  the  sides  of  a  spherical  triangle  are  arcs,  they 
can  be  measured  in  degrees.  Hence,  we  speak  of  the  side  of  a  spherical 
triangle  as  30°,  57°,  115°,  10',  etc.  In  accordance  with  this,  we  say  that 
the  limit  of  the  sum  of  the  sides  of  a  spherical  triangle  is  360°. 


30u  ELEMJ^NTARY    GEOMETRY, 


PBOrOSITXON    XVI. 

692.  Theorem. -I?! i^  ci^ni  ?f  *h  o  angles  of  a  spheri- 
cal triangle  may  be  anything  between  two  and  six  right 
angles. 

Demonstration. 

The  sum  of  the  angles  of  a  spherical  triangle  is  the  same  as  the  sum 
of  the  measures  of  the  diedrals  of  a  triedral  having'  its  vertex  at  the  centre 
of  the  sphere,  as  in  (?).  Now  the  limits  of  the  sum  of  the  measures  of 
these  diedrals  are  2  and  6  right  angles  (?).  Hence  the  sura  of  the  angles 
of  any  spherical  triangle  may  be  anything  between  2  and  6  right  angles. 

Q.  E.  D. 

6^.  Corollary. — A  spherical  triangle  may  have  one, 
two,  or  even  three  right  angles;  and,  in  fact,  it  may 
have  one,  two,  or  three  obtuse  angles;  since,  in  the 
latter  case,  the  sum^  of  the  angles  will  not  necessarily  be 
greater  than  540°. 

694.  A  Trirectaiig^ular  Spherical   Triangle   is   a 

spherical  triangle  which  has  three  right  angles. 

695.  Scholium.— It  will  be  observed  that  the  sum  of  the  angles  of  a 
spherical  triangle  is  not  constant,  as  is  the  sum  of  the  angles  of  a  plane 
triangle.  Thus,  the  sum  of  the  angles  of  a  spherical  triangle  may  be 
200°,  290°,  350°,  500°,  anything  between  180°  and  540°. 

696.  Spherical  Excess  is  the  amount  by  which  the  sum 

of  the  angles  of  a  spherical  triangle  exceeds  the  sum  of  the 
angles  of  a  plane  triangle  ;  i.  «.,  it  is  the  sum  of  the  spherical 
angles  —  180°,  or  tt. 


Exercise. — Prove  that  if  from  any  point  within  a  spherical 
triangle  arcs  of  great  circles  be  drawn  to  the  extremities  of  any 
side,  the  sum  of  these  two  arcs  is  less  than  the  sum  of  the  other 
two  sides  of  the  triangle. 


OF    THE    SPHERE,  301 


PROPOSITION    XVII. 

697.  Theorem. — The  trirectangular  triangle  is  one- 
eighth  of  the  surface  of  the  sphere. 


Demonstration. 

Pass  three  planes  through  the  centre  of  a  sphere,  respectively  perpen- 
dicular to  each  other.  They  will  divide  the  surface  into  eight  tri- 
rectangular triangles,  any  one  of  which  may  be  applied  to  any  other. 

Thus,  let  ABA'B',  ACA'C,  and  CBC'B'  be 
the  great  circles  formed  by  the  three  planes, 
mutually  perpendicular  to  each  other.  The 
planes  being  perpendicular  to  each  other,  the 
diedrals,  as  A  CO-B,  C-BO-A,  C-AO-B,  etc.,  are 
right,  and  hence  the  angles  of  the  eight  tri- 
angles formed  are  all  right. 

Also,  as  AOB  is  a  right  angle,  AB  is  a  quad- 
rant ;  as  BOC  is  a  right  angle,  CB  is  a  quadrant, 
etc.  Hence,  each  side  of  every  triangle  is  a 
quadrant. 

Whence  any  one  triangle  may  be  applied  to  any  other.  [Let  the  stu- 
dent make  the  application.] 

Hence  the  trirectangular  triangle  is  one-eighth  of  the  surface  of  the 
sphere,     q.  e.  d. 

698.  Corollary.  —  The  trirectangular  triangle  is 
equilateral  and  its  sides  are  quadrants. 


Exercise  1.  What  is  the  spherical  excess  in  a  spherical  tri- 
angle whose  angles  are  117°,  84°,  and  96°,  expressed  in  degrees  ? 
Expressed  in  right  angles  ?    Expressed  in  tt  ? 

Ans.  117°,  1A»  and  ^n. 

2.  Can  there  be  a  spherical  triangle  whose  sides  are  78°,  113°, 
and  31°  ?    Can  there  be  one  whose  sides  are  152°,  136°,  148°  ? 

3.  Can  there  be  a  spherical  triangle  whose  sides  are  52^, 
126°,  and  140°? 


ELEMENTARY    GEOMETRY, 


PROPOSITION    XVIII. 

699.  Theorem. — In  an  isosceles  spherical  triangle, 
the  angles  opposite  the  equal  sides  are  equal ;  and,  con- 
versely, //  two  angles  of  a  spherical  triangle  are  equal, 
the  triangle  is  isosceles. 


Demonstration. 
Let  ABC  be  an  isosceles  spherical  triangle,  in  which  AB  =  AC. 


Then 


angle  ABC  =  ACB. 


For,  draw  the  radii  AO,  CO,  and  BO,  form- 
ing the  edges  of  the  triedral  0-ABC. 

Now,  since  AB  =  AC,  the  facial  angles  AOB 
and  AOC  are  equal,  and  the  triedral  is  isosceles. 
Hence  the  diedrals  A-OB-C  and  A-OC-B  are  equal 
(550),  and  consequently  the  spherical  angles 
ABC  and  ACB  are  equal  (678).     Q.  e.  d. 


Fig.  311. 


Again,  if  angle  ABC  =  angle  ACB,  side  AC 
=  side  AB.     For  in   the  triedral    0-ABC,  the 

diedrals  A-OB-C  and  A-OC-B  are  equal,  whence  the  facial  angles  AOB 
and  AOC  are  equal  (550),  and  consequently  the  sides  AB  and  AC,  which 
nieasure'these  angles.     Q.  e.  d. 


700.  Corollary.— ./^W/  equilateral  spherical  triangle 
is  also  equiangular;  and,  conversely,  ^n  equiangular 
spherical  triangle  is  equilateral. 


Queries. — 1.  What  is  the  greatest  angle  which  an  equilateral 
spherical  triangle  can  have  ? 

2.  What  is  the  greatest  side  which  an  equilateral  spherical 
triangle  can  have  ? 


OF    THE    SPHERE. 


303 


PROPOSITION    XIX. 

701.  Theorem. — On  the  same  sphere,  or  on  equal 
spheres,  two  isosceles  triangles  having  two  sides  and 
the  included  angle  of  the  one  equal  to  two  sides  and 
the  included  angle  of  the  other,  each  to  each,  can  be 
superimposed,  and  are  consequently  equal. 

Demonstration. 

In  the  triangles  ABC  and  AB  C  .  let  AB  =  AC,  AB'  =  AC ;  and  let 
AB  =  AB',  BC  =  B'C,  and  angle  ABC  =  ABC  . 

Then  can  the  triangle  A  B'C  be  superimposed 
upon  ABC. 

For,  since  the  triangles  are  isosceles,  we  have 

angle  ABC  =  ACB, 

AB'C  =  ACB'  (699), 
and,  as  by  hypothesis 

ABC  =  AB'C, 

these  four  angles  are  equal,  each  to  each. 

For  a  like  reason,     AB  =  AC  =  AB'  =  AC. 

Now,  applying  AC  to  its  equal  AB,  the  extremity  A  at  A,  and  C'  at 
B,  with  the  angle  B'  on  the  same  side  of  AB  as  C,  the  convexities  of  the 
arcs  AC  and  AB  being  the  same,  and  in  the  same  direction,  the  arcs  will 
coincide.     Then,  as 

angle  ACB'  =  ABC, 

C'B'  will  take  the  direction  BC,  nnd  since  these  arcs  are  equal  by  hypoth- 
esis, B'  will  fall  at  C.  Hence  B'A  will  fall  in  CA,  as  only  one  arc  of  a 
great  circle  can  pass  between  C  and  A,  and  the  triangle  AB'C  is  super- 
imposed upon  ABC ;  wherefore  they  are  equal.    Q.  e.  d. 


Fig.  312. 


702.  Symmetrical  Spherical  Triangles  are  such  as 
have  the  parts  of  one  respectively  equal  to  the  parts  of  the  other, 
but  arranged  in  a  different  order ;  hence  such  triangles  are  not 
capable  of  superposition. 


304 


ELEMENTARY    GEOMETRY, 


Fig.  313. 


Fig.  314. 


Illustration. — In  Fig.  313,  ABC  and  A'B'C  represent  symmetrical 
spherical  triangles.     In  these  triangles, 

A  =  A',  8  =  8',  C  =  C', 

AC  =  AC,         AB  =  A'8',     and     8C  =  BC; 

nevertheless  we  cannot  conceive  one  triangle  superimposed  upon  the 
other.  Thus,  were  we  to  make  the  attempt  by  placing  A' 8'  in  its  equal 
AB,  A'  at  A,  and  B'  at  8,  the  angle  C  would  fall  on  the  opposite  side  of 
AB  from  C.  Now,  we  cannot  revolve  A'C'B'  on  AB  (or  its  chord),  and 
thus  make  the  two  coincide,  for  this  would  bring  their  convexities  to- 
gether. Nor  can  we  make  them  coincide  by  reversing  A'B'C,  and  placing 
8'  at  A,  and  A'  at  8.  For,  although  these  two  arcs  will  thus  coincide,  as 
the  angle  8'  is  not  equal  to  A,  B'C  will  not  fall  in  AC ;  and,  again,  if  it 
did,  C  would  not  fall  at  C,  since  B'C  and  AC  are  not  equal. 

But,  considering  the  triangles  ABC  and  A'B'C  in  Fig.  314,  in  which 

0  =  0', 


A 
AC 


A', 
AC, 


8  =  8', 
AB  =  A'B', 


and    80  =  B'C, 


we  can  readily  conceive  the  latter  as  superimposed  upon  the  former. 
[The  student  should  make  the  application.]  Now,  the  two  triangles  are 
equivalent  in  each  case,  as  will  subseljueiltly  appear ;  and  the  former  are 
equal.  Such  triangles  as  those  in  Fig.  313  are  called  symmetrically  equals 
while  the  latter  are  said  to  be  equal  by  superposition. 

Fig.  315  represents  the  same  triangles  as  Fig.  314,  and  exhibits  a  com- 
plete projection*  of  the  semi-circumferences  of  which  the  sides  of  the 


*  To  understand  what  is  meant  by  the  projection  of  these  lines,  conceive 
a  hemisphere  with  its  base  on  the  paper,  and  represented  by  the  circle  abc, 
and  all  the  arcs  raised  up  from  the  paper  as  they  would  be  on  the  surface  of 
such  a  hemisphere.  Thus,  considering  the  arc  a^Bb  (Fig;  315),  the  ends  a 
and  6  would  be  in  the  paper  just  where  they  are,  but  the  rest  of  the  arc 
would  be  off  the  paper,  as  though  you  could  take  hold  of  B  and  raise  it  from 


OF    THE    SPHERE. 


305 


triangles  are  arcs.  The  student  should  become  perfectly  familiar  with 
i#,  and  be  able  to  draw  it  readily.  Thus,  aAB6  is  the  projection  of  the 
semi-circumference  of  which  AB  is  an  arc,  aACc  of  the  semi-circumfer- 
ence of  which  AC  is  an  arc,  etc.,  etc. 


PROPOSITION    XX. 

703.    Theorem.  —  Two  symmetrical  spherical  tri- 
angles are  equivdlent,  i.e.,  equal  in  area. 


DEMONSTRATIOlf. 

Let  ABC  and  MB'C  be  two  symmetrical  spherical  triangles,  with  AB 
=.  AB',  AC  =  AC,  BC  :=  B'C,  A  =  A,  B  =  B  ,  and  C  =  C. 

Then  are  they  equivalent. 

Pass  circumferences  of  small  circles 
through  the  vertices  A,  B,  C,  and  A',  B',  C, 
as  abc  and  o'6V,  of  which  o  and  o'  are  the 
respective  poles. 

Now,  by  reason  of  the  mutual  equality 
of  the  sides, 

the  chord  AC  =  chord  A'C, 
chord  AB  =  chord  A'B', 
and  chord  BC  =  chord  B'C,  Fig-  3i6. 

and  as  the  small  circles  are  circumscribed  about  the  equal  plane  triangles 
ABC  and  A'B'C,  these  circles  are  equal      Hence, 

o\  =  o'^'  =oB  =  o'B'  =  oC  =  o'C\ 

and  th^  triangles  AoB  and  A'o'B',  BoC  and  B'o'C,  AoC  and  A'o'C  are 
isosceles. 

Now  call  0  the  centre  of  the  sphere,  and  draw  the  radii  OA,  OB,  DC, 
Go,  OA',  OB',  OC  ,  and  Oo'. 


the  paper  while  a  and  b  remain  fixed.  The  lines  in  the  figure  are  represen- 
tations of  lines  on  the  surface  of  such  a  hemisphere,  as  they  would  appear 
to  an  eye  situated  in  the  axis  of  the  circle  abc,  and  at  an  infinite  distance 
from  it ;  that  is,  just  as  if  each  point  in  the  lines  dropped  perpendirvlarlp 
down  upon  the  paper.  Arcs  of  great  circles  perpendicular  to  the  base  are 
projected  in  straight  lines  passing  through  the  centre,  and  oblique  arcs  are 
projected  in  ellipses.     See  Spherical  TrigonoTnetry  (97-109). 


306 


ELEMENT  A  RY    G  EOMETR  Y. 


Considering  the  triedrals  0-AoB  and 
0-AVB',  their  facial  angles  are  equal,  being 
measured  by  equal  arcs ;  hence  the  diedral 
A-oO-B  ^  A'-o'O'-B'  (0,  and  the  spherical 
angle  AoB  =  AVB'  (?).  Therefore,  the 
isosceles  triangle  AoB  =  k'o'W  (701). 

In  like  manner,  we  may  prove  the  tri- 
angles <?BC  and  o'B'C  equal,  as  also  AoC 
and  k'o'C.  

Hence,  ABC  is  equivalent  to  A'B'C,  as  '^'9-  3i6 

the  two  are  composed  of  parts  respectively  equal.     Q.  e.  d. 

If  the  poles  of  the  small  circles  fell  without  the  given  triangles,  ABC 
would  be  equivalent  to  the  sum  of  two  of  the  partial  triangles  minus  the 
third.     What  if  the  pole  fell  in  a  side? 


PROPOSITION    XXi. 

704.  Theorem.— 0«.  the  same  sphere,  or  on  equal 
spheres,  two  spherical  trian,gles  having  two  sides  and  the 
included  angle  of  the  one  equal  to  two  sides  mid  the  in- 
cluded angle  of  the  other,  each  to  each,  are  equal,  or  sym- 
metrical and  equivalent. 

DEMONSTRATrOIsr. 

Let  ABC  and  A'B'C  be  two  spherical  triangles,  having  B  =  B', 
BA  =  B'A',  and  BC  =  B'C. 

Then  are  they  either 
equal,  or  symmetrical  and 
equivalent. 

For,     passing     planes 
through  the  sides  of  each 
triangle  and  the  centre  of 
the  sphere,  two  absolutely 
or  symmetrically  equal  tri- 
edrals will  be  formed  (?).         ''»9-  3I7.  Fig.  3i8. 
Whence  the  facial  angles  AOC  and  A'OC  are  equal,  and  consequently 
AC  ^  A'C  (?).     Also,  the  diedrals  C-OA-B  and  C'OA'-B  a        -  -'   — ' 
B-OC-A  =  B'-OC'-A'  (?).     Whence  A  =  A'  and  C  =  C  (?) 

Hence  the  parts  of  ABC  are  respectively  equal  to  the  pan-  .u  A  o  0  , 
and  the  triangles  are  equal,  or  symmetrical  and  equivalent,  according  as 
the  equal  parts  are  arranged  in  the  same  or  in  a  different  oner.     t^.  jl  d, 


OF    THE    SPHERE,  307 


PROPOSITION    XXII. 

705.  Theorem. — On  the  sctme  sphere,  or  on  equal 
spheres,  two  spherical  triangles  having  two  angles  and  the 
included  side  of  the  one  equal  to  two  angles  and  the  in- 
cluded side  of  the  other,  each  to  each,  are  equal,  or  sym- 
metrical and  equivalent. 

This  is  a  direct  consequence  of  a  proposition  concerning  triedrals. 
Let  the  student  give  the  deduction. 


PROPOSITION    XXIII. 

706.  Theorem. — On  the  sam^e  sphere,  or  on  equal 
spheres,  if  two  spherical  triangles  have  two  sides  of  the  one 
equal  to  two  sides  of  the  other,  each  to  each,  and  the  in- 
cluded angles  unequal,  the  third  sides  are  unequal,  and 
the  greater  third  side  belongs  to  the  triangle  having  the 
greater  included  angle. 

Conversely,  //  the  two  sides  are  equal,  each  to  each,  and 
the  third  sides  unequal,  the  angles  included  by  the  equal 
sides  are  unequal,  and  the  greater  belongs  to  the  triangle 
having  the  greater  third  side. 

Demonstration. 

In  the  triangles  ABC  and  ABC,  let  AB  =  A'B',  AC  =  A'C,  and 
A>A. 

Then  is  BC  >  B'C 

For,  join  the  vertices  with  the  centre,  form- 
ing the  two  triedrals  0-ABC  and  0-A'B'C'. 

In  these  triedrals,  AOB  —  A' OB',  AOC  = 
A'OC,  being  measured  by  equal  arcs;  and 
C-AO-B  >  C'-A'O-B',  having  the  same  measure 
as  A  and  A'  (678).     Hence  COB  >  COB'  (?). 

Therefore  CB,  the  measure  of  COB,  >  CB', 
the  measure  of  COB'.  Fig.  3I9. 

In  like  manner,  the  same  sides  of  tVie  triangles,  and  consequently  the 
same  facial  angles  of  the  triedrals,  being  granted  equal,  and  BC  >  B'C, 
A  >  A'.  For,  BC  being  greater  than  B'C,  COB  >  COB';  whence 
B  AO-C  >  B'-A'O-C  (?),  or  A  is  greater  than  A'. 


^08  ELEMENTARY     GEOMETBY, 


PROPOSITION    XXIV. 

707.  Theorem.  —  On  the  same  sphere,  or  on  equal 
spheres,  two  spherical  triangles  having  the  sides  of  the  one 
respectively  equal  to  the  sides  of  the  other,  or  the  angles  of 
the  one  respectively  equal  to  the  angles  of  the  other,  are 
equal,  or  symmetrical  and  equivalent. 

Demonstration. 

The  sides  of  the  triangles  being  equal,  the  facial  angles  of  the  triedrals 
at  the  centre  are  equal,  whence  the  triedrals  are  equal  or  symmetrical  (?). 
Consequently,  the  angles  of  the  triangles  are  equal,  and  the  triangles  are 
equal,  or  symmetrical  and  equivalent. 

Again,  the  triangles  being  mutually  equiangular,  the  triedrals  have 
their  diedrals  mutually  equal ;  whence  the  triedrals  are  equal  or  sym- 
metrical (?).  Therefore,  the  sides  of  the  triangles  are  mutually  equal,  and 
the  triangles  are  equal,  or  symmetrical  and  equivalent.  (See  Figs.  313, 
814.) 


PROPOSITION    XXV. 

708.  Theorem. — On  spheres  of  different  radii,  mu- 
tually equiangular  triangles  are  similar  (not  equal). 

Demonstration. 

Let  ABC  and  (ibc  be  two  mutually  equiangular  spherical  triangles  on 
spheres  whose  radii  are  respectively  li  and  r,  and  let  angle  A  =  a^ 
B  =  6,  C  =  c. 

r^.      .  AB       BC      CA 

Then  is  -—=-—=  — . 

cu)        oc        ca 

For,  joining  the  vertices  of  the  triangles  with  the  centres  of  the 
spheres,  0  and  C,  the  triedrals  0-ABC  and  O'-ahc  have  their  diedrals 
mutually  equal  (?),  whence  their  facial  angles  are  mutually  equal  (?). 
Therefore  sector  AOB  is  similar  to  sector  aO%  sector  BOC  to  JO'c,  and 
sector  COA  to  cO'a. 


OF    THE    SPHERE,  309 

From  the  similarity  of  these  sectors,  we  have 

db  ~   r  ^■'^'  be  ~  r  '  ca  ~  r  ' 

AB       BC       CA 

and  hence,  —i:  =  ir  — Q.  e.  d. 

'  ab        be        ca 

709.  Scholium. — In  Spherical  Trigonometry  we  are  taught  to  find 
the  sides  of  a  splierical  triangle  having  the  angles  given.  But  in  such  a 
case  the  sides  are  found  in  degrees,  etc.,  which  does  not  determine  their 
absolute  lengths.  The  length  of  an  arc  of  any  number  of  degrees  is  not 
known  unless  the  radius  of  the  sphere  is  known. 


POLAR     OR     SUPPLEMENTAL    TRIAN- 
GLES. 

710.  One  spherical  triangle  is  Polar  to  another  when  the 
vertices  of  one  are  the  poles  of  the  sides  of  the  other,  and  the 
corresponding  vertices  lie  on  the  same  side  of  the  side  opposite. 
(For  illustration,  see  713.) 

Such  triangles  are  also  called  supplemental,  since  the  angles 
of  one  are  the  supplements  of  the  sides  opposite  in  the  other,  as 
will  appear  hereafter. 


PROPOSITION    XXVI. 

711.  Problem. — Having  a  spherical  triangle  given, 
to  draw  its  polar. 

Solution. 

Let  ABC  (Fig.  320)  be  the  given  triangle.*    From  A  as  a  pole,  with 

*  This  should  be  executed  on  a  sphere.  Few  students  get  clear  ideas  of 
polar  triangles  without  it.  Care  should  be  taken  to  construct  a  variety  of 
triangles  as  the  given  triangle,  since  the  polar  triangle  does  not  always  lie 
in  the  position  indicated  in  the  figure  here  given.  Let  the  given  triangle 
have  one  side  considerably  greater  than  90",  another  somewhat  less,  and  the 
third  quite  small.  Also,  let  each  of  the  sides  of  the  given  triangle  be 
greater  than  90°. 


310 


t:LEMENTA  R  Y     GEOMETR  Y. 


Fig.  320. 


Fig.  321. 


a  quadrant  strike  an  arc,  as  C'B'.  From  B  as  a  pole,  with  a  quadrant 
strike  the  arc  C'A';  and  from  C,  the  arc  A'B'.  Then  is  A'B'C  polar  to 
ABC. 


712.  Corollary. — If  one  triangle  is  polar  to  another, 
conversely,  the  latter  is  polar  to  the  former ;  i.  e.,  the  rela- 
tion is  reciprocal. 

Thus,  A'B'C  (Fig.  320)  being  polar  to  ABC,  reciprocally,  ABC  is  polar 
to  A'B'C ;  that  is,  A'  is  the  pole  of  CB,  B'  of  AC,  and  C  of  AB.  For 
every  point  in  A'B'  is  at^  quadrant's  distance  from  C,  and  every  point  in 
A'C  is  at  a  quadrant's  distance  from  B.  Hence,  A'  is  ai  a  quadrant's  dis- 
tance from  the  two  points  C  and  B  of  CB,  and  is  therefore  its  pole. 

[In  like  manner,  the  student  should  show  that  B'  is  the  pole  of  AC, 
and  C  of  AB.] 

713.  Scholium.— By  producing  (Fig.  331)  each  of  the  arcs  struck 
from  the  vertices  of  the  given  triangle  sufficiently,  four  new  triangles 
will  be  formed,  viz.,  A'B'C,  QC'B',  PCA',  and  RA'B'.  Only  the  first  of 
these  is  called  polar  to  the  given  triangle.  Thus,  in  A'B'C,  A',  corre- 
sponding to  A,  lies  on  the  same  side  of  CB  or  C'B'  that  A  does,  and  so  of 
any  other  corresponding  vertices. 

It  is  easy  to  observe  the  relation  of  any  of  the  parts  of  the  other  three 
triangles  to  the  parts  of  the  polar.    Thus, 

QC  =  180°  -  &', 

QB'  =  180°  -  c', 

QC'B'  =  180°  -  B'CA', 

QB'C  =  180°  — CB' A', 

and  Q  =--  A'  =  180°  -  a, 

as  will  appear  hereafter. 


OF    THE    SPHERE, 


311 


PROPOSITION    XXVII. 

714.  Theorem. — Any  angle  of  a  spherical  triangle  is 
the  supplement  of  the  side  opposite  in  its  polar  triangle  ; 
and  any  side  is  the  sicpplement  of  the  angle  opposite  in 
the  polar  triangle. 

First  Demonstration. 

Let  ABC  and  A'B'C  be  two  spherical  triangles  polar  to  each  other; 
and  let  the  sides  of  each  be  designated  as  a,  b,  c,  a',  h',  d,  a  being  op- 
posite A,  <e'  opposite  A',  h  opposite  B,  etc. 


Then 


and 


A  =  180°  -  a'^ 
P  =  180°  -  l\ 
C  =  180°  -  </, 
a  =  180°  -  A', 
b  =  180°  -  B', 
c  =  180°  -  C. 


Let  0  be  the  centre  of  the  sphere,  and 
draw  OA,  OB,  OC,  OA',  OB',  and  00'. 

The  angles  B'OA  and  B'OC  being  right 
(?),  B'O  is  perpendicular  to  the  face  AOC  (?). 

For  like  reasons,  CO  is  perpendicular  to 
the  face  AOB. 

Hence  B'OC  is  the  supplement  of  the 
diedral  B-AO-C  (512). 

But  a'  is  the  measure  of  B'OC,  and 
B-AO-C  has  the  same  measure  as  A. 

Hence,  A  =  180°  —  a'. 

In  like  manner,  we  may  show  that 

B  =  180°  -  h\      and      C  =  180°  -  d. 

Again,  since  the  edges  AO,  BO,  and  CO  are  perpendicular  to  the  faces 
R'OC,  A'OC,  and  A'OB',  we  can  show  in  like  manner  that 

a  =  180°  -  A', 

h  =  180°  -  B', 
«*"  1  e  =  180°  —  C'.     Q.  E.  D. 


Fig.  322. 


Q.  E.  D. 


312 


ELEMENTARY     GEOMETRY, 


Second  Demonsteatiok. 

Let  ABC  and  A'B'C  be  two  polar  triangles.    Let  BC,  CA,  and  AB  be 
represented  by  a,  6,  and  c  respectively,  and  B'C,  C'A',  and  AB'  by  a', 

b',  and  d. 


To  show  A  =  180°  —  a',  produce  h  and  c, 
if  necessary,  till  they  meet  the  side  a'  of  the 
triangle  polar  to  ABC  in  e  and  d. 

Now  A  is  measured  by  ed  (?).  But,  since 
^'e  -=  90°,  and  C'd  =  90°, 

B'e  +  C'd,     or     B'C  +  ed  =  180° ; 

whence,  transposing,  and  putting  a/  for  B'C, 
we  have 

ed  =  ^  =  180°  -  a' 


Fig.  323. 


In  like  manner, 
whence 
So,  also, 


C'g  +  A'f=  CA'+/^  =  180°; 
fg  =B  =  180°  -  C'A',  or  180^ 
0  =  180°  -  c'. 


To  show  that  A'  =  180°  —  a,  consider  that  A'  being  the  pole  of  CB, 
fi  is  measure  of  A'. 


whi 


Now 
ence, 


B/=90°(?),      and      Ci  =  90°; 
Bf+Ci=  180°. 


But        B/+  C^*  =fi  +  a,    wherefore   fi-\-a  =  180°  ; 
and  transposing,  and  putting  A'  for/i,  we  have  A'  =  180°  —  a. 

In  like  manner,  we  may  show  that 

B'  =  180"  -  h,      and      C  =  180°  —  e.     q.  e.  d. 
[The  student  should  give  the  details.] 

714,  a.   Corollary. — The  sum  of  the  supplements  of 
any  two  angles  of  a  spherical  triangle  is  greater  than  ^' 
supplem^ent  of  the  third  angle.     (Consider  714,  689.) 


OF    THE    SPHERE. 


313 


QUADRATURE    OF    THE    SURFACE    OF 
THE    SPHERE. 

715.  The  Quadrature*  of  a  surface  is  the  process  of  find- 
ing its  area.  The  term  is  applied  under  the  conception  that  the 
process  consists  in  finding  a  square  which  is  equivalent  to  the 
given  surface. 


PROPOSITION    XXVIII. 

716.  Ijeiiiina. — The  surf  ace  generated  hy  the  revolwtiofi 
of  a  regular  semi-polygon  of  an  even  number  of  sides, 
ahout  the^diameter  of  the  circumscribed  circle  as  an  axis, 
is  equivalent  to  the  circumference  of  the  inscribed  circle 
multiplied  by  the  axis. 

Demonstration. 

Let  ABODE  be  one-half  of  a  regular  octagon,  AE  being  the  diameter 
of  the  circumscribing  circle. 

If  the  semi-perimeter  ABODE  be  revolved  about  AE 
as  an  axis,  the  surface  generated  is  27Tr  x  AE,  r  being  the 
radius  of  the  inscribed  circle,  as  aO,  or  hO. 

This  surface  is  composed  of  the  convex  surfaces  of 
cones  and  fi-ustums  of  cones.  Thus,  AB  generates  the 
surface  of  a  cone,  BO  the  frustum  of  a  cone,  etc. 

Let  a  and  6  be  the  middle  points  of  AB  and  BO  re- 
spectively, and  draw  ow,  Be,  Z>»,  and  00  perpendicular  to 
the  axis,  and  B<Z  parallel  to  it.  Also  draw  the  radii  of  the 
inscribed  circle,  aO  and  JO.    Indicate  the  surfaces  gener-  '9*  ^  ** 

ated  by  the  sides  as  Surf.  BO,  etc.     The  areas  of  these  surfaces  are :  ^ 

Surf.  AB  =  27r  X  am  x  AB  (?),  (1) 

:^v 


Surf.  BO  =  2-  X  hn  x  BO,  etc.  (?). 


*  Latin  quadrcUus,  squared. 


14 


314 


ELEMENTA  R  Y    QEOMETR  Y. 


Now,  from  the  similar  triangles  0am  and  BAc,  we 
have 

aO  _  am 
AB  "  Ac^' 


or 


2n  X  «0  _  27rx  am 
~AB~  "  ~Ac~" ' 

Stt  X  am  X  AS  =  2Kr  x  Ac, 


whence, 
putting  r  for  aO. 

Also,  from  the  similar  triangles  Obn  and  CB^, 


we  have 


whence, 
putting  r  for  50. 


60 


bn 


BC~  Bd{=cO) 

2'iTxbO  _2nxbn 

27r  X  J»  X  BO  =  27rr  x  cO, 


Fig.  324. 


Substituting  these  values  in  (1)  and  (2),  we  obtain 


Surf.  AB  =  27rrxAc, 

Surf.  BC  =  2TTr  x  cO. 

And,  in  like  manner,         Surf.  CD  =  27rr  x  Op, 

and  Surf.  DE  =  2T:rxpE. 

Adding, 


Surf.  ABODE  =  27rr(Ac  +  eO  +  Op  +pE) 
=  27rr  X  AE. 


Finally,  since  the  same  course  of  reasoning  is  applicable  to  the  semi- 
polygons  of  16,  32,  64,  etc.,  sides,  the  truth  of  the  proposition  is  estab- 
lished. 


717-    Scholium. — This  proposition  is  only  a  particular  case  of  sur 
faces  generated  by  any  broken  line  roTolving  about  an  axis;  and  fy*p 
general  proposition  can  be  established  in  a  manner  altogether  similar  ■ 
the  method  given  above.    But  this  case  is  all  that  we  need  for  our  pres- 
ent purpose. 


OF    THE    SPHERE. 


Z\{ 


PROPOSITION    XXIX. 

718.  Theorem, — The  surface  of  a  sphere  is  equivalent 
to  four  great  circles ;  that  is,  to  ^nR^,  R  being  the  radius 
of  the  sphere. 

Demonstration. 

Let  the  semi-circumference  ABODE  revolve  upon  the  diameter  AE, 
and  thus  generate  the  surface  of  a  sphere. 

Conceive  the  half  of  a  regular  octagon  inscribed  in 
the  semicircle  ABODE;  and  let  both  the  semi-polygon 
and  the  semi-circumference  be  revolved  about  AE  as  an 
axis. 

Call  the  radius  of  the  inscribed  circle,  as  aO,  r,  and 
let  AO  =  R. 

The  surface  generated  by  the  broken  line  ABODE  is, 
by  the  last  proposition,  Stt?*  x  2R  =  AnrB. 

Now,  conceive  the  arcs  AB,  BO,  etc.,  bisected,  and  the 
chords  drawn,  and  let  r'  be  the  radius  of  the  circle  in- 


Fig.  325. 


scribed  in  the  regular  polygon  thus  formed.    The  surface  generated  by 
the  revolution  of  this  semi-polygon  is  ^ttt'R 

By  repeating  the  bisections,  the  broken  line  approximates  to  the 
semi-circumference,  the  radius  of  the  inscribed  circle  to  R,  and  the  sur- 
face generated  to  the  surface  of  the  sphere,  the  three  quantities  reaching 
their  limits  at  the  same  time.    Hence,  at  the  limit  we  have 

Surf,  of  sphere  =  2nRx2R  =  4.7rR\    q.  e.  d. 

719.  Corollary  1. — The  area  of  the  surface  of  a  sphere 
is  equivalent  to  the  circumference  of  a  great  circle  multi- 
plied by  the  diameter,  that  is,  to  2nR  x  2R,  as  above. 


720.  Corollary  2. — The  surfaces  of  spheres  are  to  each 
other  as  the  squares  of  their  radii. 

Thus,  if  R  and  R'  are  the  radii  of  two  spheres,  the  surfaces  are  47ri?* 
and  ^ttR".     Now, 


316 


ELEMENTARY    QEOMETRT. 


721.    A  Zone  is  the  portion  of  the  surface  of  a  sphere  in- 
cluded between  the  circumferences  of  two  paral- 
lel circles  of  a  sphere.     The  altitude  of  a  zone 
is  the  distance    between   the    parallel    circles 
whose  circumferences  form  its  bases. 


Illustration. — The  surface  generated  by  CB,  or 
any  arc  of  the  circle  ABODE,  etc..  as  the  semicircle 
revolves  about  AE  as  an  axis,  conforms  to  the  defini- 
tion, and  is  a  zone.  Such  a  portion  of  the  surface  as  is 
generated  by  AB  is  called  a  zone  with  one  hase,  the 
circle  whose  circumference  would  form  the  upper  base 
having  become  tangent  to  the  sphere.  The  altitude  of 
the  zone  generated  by  CB  is  a5,  and  of  that  generated  by  AB  the  alti- 
tude is  Aa. 


Fig.  32G. 


PROPOSITION    XXX. 

722.  Theorem. — The  area  of  a  zone  is  equal  to  27TaR, 
a  being  the  altitude  of  the  zone  and  R  the  radius  of  the 
sphere. 

Demonstration. 

It  is  evident  that  in  passing  to  the  limit,  the  surface  generated  by 
such  a  portion  of  the  broken  line  as  lies  between  C  and  B,  Fig.  326,  is 
measured  by  the  circumference  of  the  inscribed  circle  multiplied  by  db. 
Hence,  at  the  limit,  the  zone  generated  by  arc  BC  is  measured  by 

2rrifxaJ,    that  is,     3n-ai?, 
representing  db  by  a.    Q.  e.  d. 

723.  Corollary. — On  the  same  sphere,  or  on  equal 
spheres,  zones  are  to  each  other  as  their  altitudes,  and  any 
zone  is  to  the  surface  of  the  sphere  as  the  altitude  of  the 
zone  is  to  the  diameter  of  the  sphere. 


OF    THE    SPHERE. 


ai7 


OF      LU  N  ES. 

724.  A  Ijuiie  is  a  portion  of  the  surface  of  a  sphere- included 
by  two  semi-circumferences  of  great  circles. 

The  surface  km^n  is  a  lune. 

725.  The  Angle  of  the   Lune  is 

the  angle  included  by  the  arcs  which  form      \ 
its  sides  ;   or,  what  is  the  same  thing,  the 
measure  of  the  diedral  included  between  the 
great  circles. 

Thus,  the  spherical  angle  wAn,  or  the  measure  of  the  diedral  w-AB-n 
is  the  augle  of  the  lune  A^nB/i. 

726.  An  Ungula,  or  Spherical  Wedge,  is  that  portion 
of  a  sphere  included  between  two  semi-great-circles,  as  AwB  and 
knB.  It  has  a  lune  for  its  convex  surface  and  a  diameter  for  its 
edge. 


PROPOSITION    XXXI. 

727.  Theorem. — On    the   same    sphere,  or  on   equal 
spheres,  lunes  which  have  equal  angles  are  equal. 

Demonstration. 

[This  is  readily  effected  by  applying  one  to  the  other.     Let  the  stu- 
dent make  the  application.] 


Exercise. — Can  there  be  a  spherical  triangle  whose  angles  are  152°,- 
136°,  and  148°?  One  whose  angles  are  152°,  136°,  and  168'/  (See 
714,  a.) 


318 


ELEMENTARY    GEOMETRY. 


PROPOSITION    XXXII. 

728.  Theorem. — The  area  of  a  lune  is  to  the  area  of 
the  surface  of  the  sphere  on  which  it  is  situated  as  the 
angle  of  the  lune  is  to  four  right  angles. 


First  Demon^stration. 

Let  S  represent  the  area  of  the  surface  of  the  sphere  generated  by 
the  revolution  of  the  semicircle  IV1AN  about  MN  as  an  axis,  and  L  the 
area  of  the  lune  whose  angle  is  AMD,  or  AOD. 


Then  is    —  = 


AOD 


4  riglit  angles 

In  the  generation  of  8  and  L  by  the 
semi-circumference  MAN,  the  middle 
point,  A,  of  the  semi-circumference  gen- 
erates the  great  circle  ACDBF,  on  which 
the  angles  of  the  lunes  are  measured  (?). 

Now  A  generates  equal  and  coinci- 
dent parts  of  arc  AD  and  circumference 
ACDBFA,  in  the  same  time  that  MAN 
generates  corresponding  equal  and  co- 
incident parts  of  L  and  8. 

arc  AD 


Fig.  328. 


Hence,  if 


circf.  ACDBF  ""£' 


and 


arc  AD 


AOD 


circf.  ACDBF       4  right  angles 


(?).     Q.  E.  D. 


Second  Demonstration. 

Let  S  represent  the  area  of  the  surface  of  the  sphere  generated  by 
the  revolution  of  the  semicircle  MAN  about  MN  as  an  axis,  and  L  the 
area  of  the  lune  whose  angle  is  AMD^  or  AOD. 

Now  the  angles  AOD  and  the  sum  of  the  four  right  angles  AOD, DOB^ 
BOF,  FOA  are  at  least  commensurable  by  an  infinitesimal  unit.    Let  i  be 


OF    THE    SPHERE. 


319 


their  common  measure,  and  let  it  be  contained  in  AOD  n  times,  and* in 
the  four  right  angles  m  times,  so  that 

AOD  ^  n 

4  right  angles  ~"  m 


Now  conceive  the  circumference  divided  into  m  equal  parts,  and 
radii  drawn  to  the  points  of  division ;  and  through  their  extremities  let 
serai-circumferences  be  drawn.  Then  is  L  divided  into  n  lunes,  each 
equal  to  one  of  the  m  equal  lunes  into  which  S  is  divided  (727),  so  that 


Hence, 


L  _n 

8      m 

L 

AOD 

8  ~ 

4  right  angles 

Q.  E.  D. 


Third  Demonstration. 

Let  5  be  the  surface  of  the  sphere,  and  ACEB  =  X  be  a  lune  whose 
angle  is  the  spherical  angle  CAB,  or  what  is  the  same  thing,  the  plane 
angle  BOC  measured  by  the  arc  CB,  of  which  A  is  the  pole. 


Then  is        7;  = 


CAB 


4  right  angles 

For,  first,  suppose  the  arc  CB  commensurable 
with  the  circumference  BCwDra,  and  suppose  that 
they  are  to  each  other  as  5  :  24. 

Divide  CB  into  five  equal  arcs,  and  the  entire 
circumference  BCmDn  into  twenty-four  arcs  of  the 
same  length,  and  pass  arcs  of  great  circles  through 
A  and  these  points  of  division.     Thus  the  lune  is  ^'9-  ^29. 

divided  into  five  equal  lunes,  and  the  entire  surface  into  twenty-four 
equal  lunes  of  the  same  size.    These  lunes  are  equal  to  each  other  (727). 


Hence, 
Now, 


COB 


4  right  angles 


24" 

CB 

BCmDw. 


_5^ 
24' 


Therefore 


L  _  COB  (or  CAB) 

8  ~  ^  right  angles 


%  E.  D, 


320  ELEMENTARY    GEOMETRY. 

•  If  the  angle  of  the  lune  is  incommensurable  with  four  right  angles, 
or,  what  is  the  same  thing,  if  the  arc  BC  is  not  commensurable  with  the 
circumference,  let  us  assume 

-  =  -?^  (1) 

in  which  BL  <  BC. 

Conceive  the  circumference  BCmDTi  divided 
into  equal  parts,  each  of  which  is  less  than  CL, 
the  assumed  difference  between  BC  and  BL. 
Then  conceive  one  of  these  equal  parts  applied 
to  BC  as  a  measure,  beginning  at  B.  Since  the 
measure  is  less  than  LC,  one  point  of  division,  at  ^'9'  ^^°' 

least,  will  fall  between  L  and  C.    Let  I  be  such  a  point,  and  pass  the  arc 
of  a  great  circle  through  A  and  I. 

XT  lune  AIEB  Bl 

8  BCmDri  ^  ' 

since  the  arc  Bl  is  commensurable  with  the  circumference.     In  (1)  and 
(2),  the  consequents  being  equal,  the  antecedents  should  be  proportional ; 

L  BL 


hence  we  should  have 


lune  AIEB       Bl 


But  this  is  absurd,  since  lune  ACEB  >  lune  AIEB,  whereas  BL  <  Bl, 
that  is,  an  improper  fraction  equals  a  proper  fraction. 

In  a  similar  manner,  we  may  reduce  the  assumption  to  an  absurdity, 
if  we  assume  BL  >  BC. 

L  BC 

Hence,  as  the  ratio  of  -^  can  neither  be  greater  nor  less  than  , 

it  is  equal  thereto,  and 

L  BC  BOC 


8    '  EQrnXin      4  right  angles 


Q.  E.  D. 


^ 


729.  Scholium. —  To  obtain  the  ourea  of  a  lune  whose  angle  is  known, 
find  the  area  of  the  sphere,  and  multiply  it  by  the  ratio  of  the  angle  of 
the  lune  (in  degrees)  to  360°.  Thus,  B  being  the  radius  of  the  sphere, 
4i7rB^  is  the  surface  of  the  sphere ;  and  the  lune  whose  angle  is  30°  is  ^jj 
or  Y^2^  the  surface  of  the  sphere,  i.  e.,  ^^  oi  A^nB^  =  ^tzB^. 

730.  Corollary. — The  sum  of  several  lunes  on  the  same 
sphere  is  equal  to  a  lune  whose  angle  is  the  sum  of  the 
angles  of  the  lunes ;  and  the  difference  of  two  lunes  is  a 
lune  whose  angle  is  the  difference  of  their  angles. 


OF    THE    SPHERE. 


321 


731.  Corollary. — Ungulas  hear  the  same  ratio  to  the 
volume  of  the  sphere  that  the  corresponding  lunes  do  to 
the  area  of  the  surface. 


PROPOSITION    XXXIII. 

732.  Theorem. — If  two  semi-circumferences  of  great 
circles  intersect  on  the  surface  of  a  hemisphere,  the  sum 
of  the  two  opposite  triangles  thus  formed  is  equivalent  to 
a  lune  whose  angle  is  that  included  by  the  semi-circum- 
ferences. 

Demonstration. 

Let  the  semi-circumferences  CEB  and  DEA  intersect  at  E  on  the  sur- 
face of  the  hemisphere  whose  base  is  CABD. 

Then  the  sum  of  the  triangles  CED  and 
AEB  is  equivalent  to  a  lune  whose  angle  is 
AEB. 

For,  let  the  semi-circumferences  CEB  and 
DEA  be  produced  around  the  sphere,  inter- 
secting on  the  opposite  hemisphere,  at  the 
extremity  F  of  the  diameter  through  E. 

Now,  FBEA  is  a  lune  whose  angle  is  AEB. 

Moreover,  the  triangle  AFB  is  equivalent 
to  the  triangle  DEC ;  since  F'g-  33'- 

angle  AFB  =  AEB  =  DEC, 

side  AF  =  side  ED, 

each  being  the  supplement  of  AE ;  and 

BF  =  CE, 

each  being  the  supplement  of  EB. 

Hence,  the  sum  of  the  triangles  CED  and  AEB  is  equivalent  to  the 
lune  FBEA.     Q.  e.  d. 


ELEMENTARY    GEOMETRY. 


PROPOSITION    XXXIV. 

733.  Theorem. — The  area  of  a  spherical  triangle  is  to 
the  area  of  the  surface  of  the  hemisphere  on  which  it  is 
situated,  as  its  spherical  excess  is  to  four  right  angles,  or 
360°. 

Demonstration. 

Let  ABC  be  a  spherical  triangle  whose 
angles  are  represented  by  A,  B,  and  C;  let  T 
represent  the  area  of  the  triangle,  and  H  the 
area  of  the  surface  of  the  hemisphere. 

T  _  A  +  B  +  C-180° 
R  ~  360° 


Then  is 


Let  lune  A  represent  the  lune  whose  angle  is 
the  angle  A  of  the  triangle,  i.  e.,  angle  CAB,  and 
in  like  manner  understand  lune  B  and  lune  C. 


Fig.  332. 


Now,  triangle  AHG  +  AED  =  lune  A  (732), 

BHI  +  BEF  =  lune  B, 

CGF  +  CDI    =  lune  C. 

Adding,  3ABC  +  hemisphere  =  lune  (A  +  B  +  C)* 


(1) 


by  (730),  and  since  the  six  triangles  AHG,  AED,  BHI,  BEF,  CGF,  and  CDI 

make  the  whole  hemisphere  and  2ABC  besides,  ABC  being  reckoned  three 
times 

From  (1)  we  have,  by  transposing,  and  remembering  that  a  hemi- 
sphere is  a  lune  whose  angle  is  180°  (730),  and  dividing  by  2, 

ABC  =  i  lune  (A  +  B  +  C  -  180°).t 


But,  by  (728), 

i lune  (A 

+  B  + 
H 

C- 

-  180°) 

A  +  B 

+  C- 

360° 

180° 

Therefore, 

T 

H  ~ 

.  A 

+  B  +  C 
•    360^ 

-  180° 

Q.  E, 

D. 

*  This  signifies  the  lun<^whose  angle  is  A  -f  B  +  C,  which  is  of  course 
the  sum  of  the  three  lunes  whose  angles  are  A,  B.  and  C. 

f  This  signifies  one-half  the  lune  whose  angle  is  A  +  B  +  C— 180°. 


OF    THE    SPHERE.  323 

734.  Scholium  1. —  To  find  the  area  of  a  spherical  triangle  on  a  given 
sphere,  the  angles  of  the  triangle  being  given,  we  have  simply  to  multiply 
the  area  of  the  hemisphere,  *.  e. ,  2n-i2^,  by  the  ratio  of  the  spherical  excess 
to  360".     Thus,  if  the  angles  are 

A  =  110°,        B  =  80°,    and    C  =  50°,  j  s^ 

we  have  ^"^ 

area  ABC  =  a.if'  x  ^^L?4^=J^1  =  2riP  x  ^  =  ^nS?. 


735.  Scholium  3. — This  proposition  is  often  stated  thus:  Tlie  area 
of  a  spherical  triangle  is  equal  to  its  spherical  excess  multiplied  by  the  trirec- 
tangular  triangle.  When  so  stated,  the  spherical  excess  is  to  be  estimated 
in  terms  of  the  right  angle ;  i.  e.,  having  subtracted  180^  from  the  sum  of 
its  angles,  we  are  to  divide  the  remainder  by  90°,  thus  getting  the  spheri- 
cal excess  in  right  angles.  In  the  example  in  the  preceding  scholium, 
the  spherical  excess  estimated  in  this  way  would  be 

110°  +  80^  +  50^  -  180°  _  2 
90°  ~3 

and  the  area  of  the  triangle  would  be  |  of  the  trirectangular  triangle. 
Now,  the  trirectangular  triangle  being  |-  of  the  surface  of  the  sphere  (?), 
is  ^  of  47ri?',  or  ^7ri^^  This  multiplied  by  |  gives  ^Tri?*,  the  same  as 
above. 

The  proportion 

ABC  ^  A  +  B  +  C  -  180° 

surf,  of  hemisph.  360'  ' 

is  readily  put  into  a  form  which  agrees  with  the  enunciation  as  given  in 
this  scholium.     Thus, 

surf,  of  hemisph.  =  27ri?» ; 
whence, 

ARP  -c)^m  ^  A  +  B  +  C-180^       ,    _- .      A+B  +  C-180° 
ABC  -  2^i?  X ^— =  ^.m  X 


324 


ELEMENTARY    GEOMETRY. 


VOLUME    OF    SPHERE. 


PROPOSITION    XXXV. 

736.  Theorem. — The  volume  of  a  sphere  is  equal  to 
the  area  of  its  surface  multiplied  by  one-third  of  the  ra- 
dius, that  is,  f  7Ti23j  R  being  the  radius. 


Demonstration. 

Let  OL  =  i^  be  the  radius  of  a  sphere. 

Conceive  a  circumscribed  cube,  that  is,  a 
cube  wliose  faces  are  tangent  planes  to  the 
sphere.  Draw  lines  from  the  vertices  of  each  of 
the  polyedral  angles  of  the  cube  to  the  centre 
of  the  sphere,  as  AO,  BO,  DO,  CO,  etc.  These 
lines  are  the  edges  of  six  pyramids,  having  for 
their  bases  the  faces  of  the  cube,  and  for  a  com- 
mon altitude  the  radius  of  the  sphere  (?).  Hence  ^^' 
the  volume  of  the  circumscribed  cube  is  equal  to  its  surface  multiplied 
by  ^B. 

Again,  conceive  each  of  the  triedral  angles  of  the  cube  truncated  by 
planes  tangent  to  the  sphere.  A  new  circumscribed  solid  will  thus  be 
formed,  whose  volume  will  be  nearer  that  of  the  sphere  than  is  that  of  the 
circumscribed  cube.  Let  a5c  represent  one  of  the  tangent  planes.  Draw 
from  the  polyedral  angles  of  this  new  solid,  lines  to  the  centre  of  the 
sphere,  as^^O.  50,  and  cO,  etc. ;  these  lines  will  form  the  edges  of  a  set  of 
pyramids  whose  bases  constitute  the  surface  of  the  solid,  and  whose  com- 
mon altitude  is  the  radius  of  the  sphere  (?).  Hence  the  volume  of  this 
solid  is  equal  to  the  product  of  its  surface  (the  sum  of  the  bases  of  the 
pyramids)  into  \R. 

Now,  this  process  of  truncating  the  angles  by  tangent  planes  may  be 
conceived  as  continued  indefinitely ;  and,  to  whatever  extent  it  is  carried, 
it  will  always  be  true  that  the  volume  of  the  solid  is  equal  to  its  surface 
multiplied  by  ^R.  Therefore,  as  the  sphere  is  the  limit  of  this  circum- 
scribed solid,  we  have  the  volume  of  the  sphere  equal  to  the  surface  of 
the  sphere,  which  is  47ri2'  multiplied  by  |i2,  i.  e.,  to  ^nW.     q,.  e.  d. 


OF    THE    SPHERE.  325 

737.  CoEOLLARY. — The  surface  of  the  sphere  may  he 
conceived  as  consisting  of  an  infinite  nuinher  of  infinitely 
small  plane  faces,  and  the  volume  as  composed  of  an  infi- 
nite number  of  pyramids  having  these  faces  for  their 
bases,  and  their  vertices  at  the  centre  of  the  sphere,  the 
common  altitude  of  the  pyramids  being  the  radius  of  the 
sphere. 


738.  A  Spherical  Sector  is  a  portion  of  a  sphere  gener- 
ated by  the  revolution  of  a  circular  sector  about  the  diameter 
around  which  the  semicircle  which  generates  the  sphere  is  con- 
ceived to  revolve.  It  has  a  zone  for  its  base  ;  and  it  may  have 
as  its  other  surfaces  one  or  two  conical  surfaces,  or  one  conical 
and  one  plane  surface. 

Illustration. — Thus,  let  ab  be  the  diam- 
eter around  which  the  semicircle  aEl  revolves 
to  generate  the  sphere.  The  solid  generated 
by  the  circular  sector  AOB  will  be  a  spherical 
sector  ha^^ng  the  zone  generated  by  AB  for  its 
base;  and  for  its  other  surface,  the  conical  sur- 
face generated  by  AO.  The  spherical  sector 
generated  by  COD  has  the  zone  generated  by 
CD  for  its  base;  and  for  its  other  surfaces, 
the  concave  conical  surface  generated  by  DO, 
and  the  convex  conical  surface  generated  by  CO.  The  spherical  sector 
generated  by  EOF  has  the  zone  generated  by  EF  for  its  base,  the  plane 
generated  by  EO  for  one  surface,  and  the  concave  conical  surface  gener- 
ated by  FO  for  the  other.  ^  '  ., 

739.  A  Spherical  Segrment  is  a  portion  of  the  sphere 

included  by  two  parallel  planes,  it  being  understood  that  one  of 
the  planes  may  become  a  tangent  plane.  In  the  latter  case,  the 
segment  has  but  one  base  ;  in  other  cases,  it  has  two.  A  spheri- 
cal segment  is  bounded  by  a  zone  and  one,  or  two,  plane  surfaces. 


326 


ELEMENTARY    GEOMETRY. 


PROPOSITION    XXXVI. 

740.  Theorem. — The  volume  of  a  spherical  sector  is 
equal  to  the  product  of  the  zone  which  forms  its  base  into 
one-third  the  radius  of  the  sphere. 

Demonstration. 

A  spherical  sector,  like  the  sphere  itself,  may  be  conceived  as  consist- 
ing of  an  infinite  number  of  pyramids  whose  bases  make  up  the  base  of  the 
sector,  and  whose  common  altitude  is  the  radius  of  the  sphere.  Hence, 
the  volume  of  the  sector  is  equal  to  the  sum  of  the  bases  of  these  pyra- 
mids, that  is,  the  surface  of  the  sector,  multiplied  by  one-third  their  com- 
mon altitude,  which  is  one-third  the  radius  of  the  sphere,     q.  e.  d. 

741.  Corollary. — The  volumes  of  spherical  sectors  of 
the  same  sphere,  or  of  equal  spheres,  are  to  each  other  as 
the  zones  which  form  their  bases ;  and,  since  these  zones  are 
to  each  other  as  their  altitudes  (723),  the  sectors  are  to  each 
other  as  the  altitudes  of  tJie  zones  luhich  form>  their  bases. 


PROPOSITION    XXXVII. 

742.  Theorem.— 17^6  volume  of  a  spherical  segment 
of  one  base  is  ttA^{R  —  \A),  A  being  the  altitude  of  the 
segment,  and  R  the  radius  of  the  sphere. 

Demonstration. 
Let  AG  =  JB,  and  CD  =  A. 

Then  is  the  volume  of  the  spherical  segment 
generated  by  the  revolution  of  ACD  about  CO 
equal  to  ttA^  {R  —  ^A). 

For,  the  volume  of  the  spherical  sector  gener- 
ated by  AOC  i^  the  zone  generated  by  AC,  multi- 
plied by  \B,  or  ^TTAIty.^R  =  ^t:AR\  From 
this  we  must  subtract  the  cone,  the  radius  of 
whose  base  is  AD,  and  whose  altitude  is  DO. 


Fig.  335. 


OF    THE    SPHERE, 

To  obtain  this,  we  have 

DO  =  R-A\ 

whence,  from  the  right-angled  triangle  ADO, 


327 


AD  =  V^  -{E-  Af  =  ^2AB  -  A\ 

Now,  the  volume  of  this  cone  is  ^OD  x  ttAD^  or 

^TT  (B  —  A)  i2AB  -  A"")  =  ^n  {2AR'  —  3A'B  +  A^). 

Subtracting  this  from  the  volume  of  the  spherical  sector,  we  have 
^kAB^  -  ^TT  (2^if  -  dA^'B  +  A')  =  n  (A'R  -  ^A^ 

=  nA^B-^A).      Q.  E.  D. 


743.  Scholium. — The  volume  of  a  spherical 
segment  with  two  bases  is  readily  obtained  by 
taking  the  difference  between  two  segments  of 
one  base  each.  Thus,  to  obtain  the  volumes  of 
the  segment  generated  by  the  revolution  of  JCAc 
about  aO,  take  the  difference  of  the  segments 
whose  altitudes  are  ac  and  ab. 


Fifl.  336. 


SPHERICAL    POLYGONS    AND  SPHERI- 
CAL   PYRAMIDS. 

744.  A  Spherical  Polygon  is  a  portion  of  the  surface  of 
a  sphere  bounded  by  several  arcs  of  great  circles. 

745.  The  Diagonal  of  a  spherical  polygon  is  an  arc  of  a 
great  circle  joining  any  two  non-adjacent  vertices. 

746.  A  Spherical  Pyramid  is  a  portion  of  a  sphere  hav- 
ing for  its  base  a  spherical  polygon,  and  for  its  lateral  faces  the 
circular  sectors  formed  by  joining  the  vertices  of  the  polygon 
with  the  centre  of  the  sphere. 


328  ELEMENTARY    GEOMETRY. 

747.  The  elementary  properties  of  spherical  polygons  and 
spherical  pyramids  are  so  readily  deduced  from  the  corresponding 
properties  of  polyedral  angles,  spherical  triangles,  etc.,  that  we 
leave  them  for  the  pupil  to  demonstrate,  merely  stating  a  few 
fundamental  theorems. 

748.  Theorem.— ^6  angles  of  a  spherical  polygon 
and  its  sides  sustain  the  same  general  relations  to  each 
other  as  the  diedral  and  facial  angles  of  a  polyedral 
angle  having  for  its  edges  the  radii  of  the  sphere  drawn 
to  the  vertices  of  the  polygon. 

749.  Theorem. — T^^e  sum  of  the  sides  of  a  convex 
spherical  polygon  may  be  anything  between  0°  and  360°. 

760.  Theorem. — The  sum  of  the  angles  of  a  spherical 
polygon  may  be  anything  between  ^n  —  4  and  Qn  — 12 
right  angles,  n  being  the  number  of  sides. 

751.  The  Spherical  Excess  of  a  spherical  polygon  is  the 
excess  of  the  sum  of  its  angles  over  the  sum  of  the  angles  of  a 
plane  polygon  of  the  same  number  of  sides. 

752.  Theorem. — The  spherical  excess  of  a  spherical 
polygon  of  n  sides,  the  sum  of  whose  angles  is  S,  is 

^+ 360° -71.180°. 

753.  Theorem. — The  area  of  a  spherical  polygon  is  to 
the  area  of  the  surface  of  the  hemisphere  on  which  it  is 
situated  as  its  spherical  excess  is  to  four  right  angles. 

754.  Theorem.— IT^e  volume  of  a  spherical  pyramid 
is  the  area  of  its  base  multiplied  by  one-third  the  radius 
of  the  sphere  on  which  it  is  situated. 


EXERCISES,  329 


EXERCISES. 

755.  1.  What  is  the  circumference  of  a  small  circle  of  a  sphere 
whose  diameter  is  10,  the  circle  being  at  3  from  the  centre  ? 

Ans.  25.1328. 

2.  Construct  on  the  spherical  blackboard  a  spherical  angle  of 
60°.    Of  45°.     Of  90°.     Of  120°.     Of  250°. 

Suggestions.— Let  P  be  the  point  where  the  vertex  of  the  required 
angle  is  to  be  situated.  With  a  quadrant  strike  an  arc  passing  through 
P,  which  shall  represent  one  side  of  the  required  angle.  From  P  as  a 
pole,  with  a  quadrant  strike  an  arc  from  the  side  before  drawn,  which 
shall  measure  the  required  angle.  On  this  last  arc  lay  off  from  the  firet 
side  the  measure  of  the  required  angle,*  as  60°,  45%  etc.  Through  the 
extremity  of  this  arc  and  P  pass  a  great  circle  (?). 

3.  On  the  spherical  blackboard  construct  a  spherical  triangle 
ABC,  having  AB  =  100°,  AC  =  80°,  and  A  =  58°. 

4.  Construct  as  above  a  spherical  triangle  ABC,  having  AB  = 
75°,  A  =.  110°,  and  B  =  87°. 

5.  Construct  as  above,  having  AB  =  150°,  BC  =  80°,  and 
AC  =  100°.     Also  having  AB  =  160°,  AC  =  50°,  and  CB  =  85°. 

6.  Construct  as  above,  having  A  =  52°,  AC  =  47°,  and  CB 
=  40°. 

Suggestions. — Construct  the  angle  A  as  before  taught,  and  lay  off 
AC  from  A  equal  to  47'',  with  the  tape.  This  determines  the  vertex  C. 
From  C,  as  a  pole,  with  an  arc  of  40°,  describe  an  arc  of  a  small  circle; 
in  this  case  this  arc  will  cut  the  opposite  side  of  the  angle  A  in  two 
places.  Call  these  points  B  and  B'.  Pass  circumferences  of  great  circles 
through  C,  and  B,  and  B'.     There  are  two  triangles,  ACB  and  ACB'. 

7.  Construct  on  the  spherical  blackboard  a  spherical  triangle 
ABC,  having  A  =  59°,  AC  =  120°,  and  AB  =  88°. 


♦  For  this  purpose,  a  tape  equal  in  length  to  a  semi-circumference  of  a 
jrreat  circle  of  the  sphere  used,  and  marked  off  into  180  equal  parts,  will  be 
convenient.     A  strip  of  paper  may  be  used. 


330  ELEMENTARY    GEOMETRY. 

8.  Construct  a  triangle  whose  angles  are  160°,  150°,  and  140°. 

9.  Can  there  be  a  spherical  triangle  whose  angles  are  85°, 
120°,  and  150°  ?  Try  to  construct  such  a  triangle  by  first  con- 
structing its  polar. 

10.  What  is  the  area  of  a  spherical  triangle  on  the  surface  of  a 
sphere  whose  radius  is  10,  the  angles  of  the  triangle  being  85°, 
120°,  and  110° ?  Ans.    235.6-j-. 

11.  What  is  the  area  of  a  spherical  triangle  on  a  sphere  whose 
diameter  is  12,  the  angles  of  the  triangle  being  82"",  98°,  and 
100°? 

12.  A  sphere  is  cut  by  five  parallel  planes  at  7  from  each  other. 
What  are  the  relative  areas  of  the  zones  ?     What  of  the  segments  ? 

13.  Considering  the  earth  as  a  sphere,  its  radius  would  be 
3958  miles,  and  the  altitudes  of  the  zones.  North  torrid  =  1578, 
North  temperate  =  2052,  and  North  frigid  =  328  miles.  What 
are  the  relative  areas  of  the  several  zones  ? 

Suggestion. — The  student  should  be  careful  to  discriminate  between 
the  width  of  a  zone  and  its  altitude.  The  altitudes  are  found  from  their 
widths,  as  usually  given  in  degrees,  by  means  of  Trigonometry. 

14.  The  earth  being  regarded  as  a  sphere  whose  radius  is 
3958  miles,  what  is  the  area  of  a  spherical  triangle  on  its  surface, 
the  angles  being  120°,  130°,  and  150°  ?  What  is  the  area  of  a 
trirectangular  triangle  on  the  earth's  surface  ? 

15.  In  the  spherical  triangle  ABC,  given  A  =  58°,  B  =  67°, 
and  AC  =  81°  ;  what  can  you  afiirm  of  the  polar  triangle  ? 

16.  What  is  the  volume  of  a  globe  which  is  2  feet  in  diameter  ? 
What  of  a  segment  of  the  same  globe  included  by  two  parallel 
planes,  one  at  3  and  the  other  at  9  inches  from  the  centre,  the 
centre  of  the  sphere  being  without  the  segment?  What  if  the 
centre  is  within  the  segment? 

17.  Compare  the  convex  surfaces  of  a  sphere  and  its  circum- 
scribed cylinder. 

18.  Compare  the  volumes  of  a  sphere  and  its  circumscribed 
cube,  cylinder,  and  cone,  the  vertical  angle  of  the  cone  being  60°. 

19.  If  a  and  b  represent  the  distances  from  the  centre  of  a 
sphere  whose  radius  is  r,  to  the  bases  of  a  spherical  segment,  show 
that  the  volume  of  the  segment  is  tt  [r^  (b  —  a)  —  \{h^  —  a^)]. 


THE     INFINITESIMAL     METHOD. 

The  author  is  a  firm  believer  in  both  the  logical  soundness  and  the 
practical  advantages  of  the  strict  infinitesimal  method.  Hence  he  has  intro- 
duced it — though  generally  as  an  alternative  method — in  those  cases  in 
which  the  incommensurability  of  geometrical  magnitudes  by  a  finite  unit 
makes  the  old  demonstrations  cumbrous. 

As  to  the  logical  soundness  of  the  method,  he  has  not  the  shadow  of  a 
doubt.  The  well-known  logical  principle,  that,  if  we  create  a  certain  cate- 
gory of  concepts,  under  certain  definite  laws,  use  them  in  our  argument  in 
accordance  with  these  laws,  and  finally  eliminate  them,  the  argument  being 
conducted  according  to  correct  logical  principles,  the  final  results  are 
correct,  covers  the  entire  case.  Now  the  two  essential  laws  of  infinitesimals 
are,  (1)  Infinitesimals  of  the  same  order  have  the  same  relations  among 
themselves  as  finite  quantities ;  and  (2)  Infinitesimals  in  comparison  with 
finites,  are  zero. 

But  the  simple  exposition  given  in  the  text  (340-342)  is  quite  adequate 
to  show  that  the  method  can   introduce  no  conceivable  error.     Thus,  if 

—  —  <r,  all  the  quantities  being  finite,  and  if  i  is  an  infinitesimal,  =  a 

must  be  true,  and  i  must  be  0  in  the  relation.     Otherwise  solving  the  equa- 

tion  we  have  i  =  an—m,  a  finite  quantity,  unless  a  —  —- 

Of  the  immense  practical  utility  of  the  method  there  can  be  no  question. 
All,  from  Lagrange  down,  have  acknowledged  it.  I  know  of  no  extended 
treatise  which  does  not  in  some  way  imply  it.  Why,  then,  should  not  the 
pupil  become  familiar  with  it  early  in  his  course  ? 

As  to  the  method  of  limits  it  is  not  at  all  difficult  to  show  that  it  is 
identical  with  the  infinitesimal  method,  in  its  fundamental  principles. 
Moreover,  there  is  a  sort  of  jugglery  in  the  very  first  step  in  the  method  of 
limits  which  quite  transcends  any  difficulty  that  the  method  of  infinitesimals 
presents.  Thus,  we  give  the  variable  an  increment,  assume  that  the  func- 
tion takes  a  related  increment,  manipulate  the  function,  and  then  make  the 
increment  of  the  variable  zero  (whence  the  increment  of  the  function 
becomes  zero),  and,  presto,  we  have  a  finite  relation  between  two  zeros! 
And  this  is  the  "  simple "  fundamental  conception  which  the  tyro  is  sup- 
posed to  see  at  a  glance ! 


332  APPENDIX. 


NOTE    ON    (182),   (343),   (587),   (628),  AND   (728). 

These  propositions  are  of  a  class  in  wliich  the  incommensurability  by  a 
finite  unit  of  certain  lines  introduces  particular  diflficulty,  which  difficulty 
disappears  at  once  if  we  admit,  as  in  the  infinitesimal  theory,  that  these  lines 
are  commensurable  by  an  infinitesimal  unit.  Also,  by  the  introduction  of 
the  principle  of  the  generation  of  one  magnitude  by  the  motion  of  another, 
very  simple  demonstrations  are  afforded. 

In  the  text  the  author  has  given  illustrations  of  the  three  sorts  of 
demonstrations.  In  (182)  we  have  the  old  method  of  avoiding  the  difficulty 
which  grows  out  of  the  incommensurability,  by  the  reductio  absurdum. 
The  objection  to  this  is  not  any  objection  to  the  reductio  absurdum  as  a 
method  of  reasoning.  But  why  use  so  cumbrous  a  method,  when  other 
exceedingly  simple  methods  are  at  hand,  and  methods  involving  principles 
so  necessary  to  subsequent  use  ? 

In  (728)  tlie  three  methods  are  given.  In  (343)  and  (628)  the  methods 
involving  generation  by  motion,  and  the  infinitesimal  method,  are  given. 


NOTE    ON    (182). 

1.  To  prove  this  proposition  by  means  of  the  conception  of  the  genera- 
tion of  magnitudes  by  the  motion  of  other  magnitudes,  we  do  not  need  the 

AOB 

Lemma.     Thus,  referring  to  Fig.  85,  p.  89,  we  are  to  prove  that  — -  = 

arc  AB 
arc  DE 

Let  the  sector  AOB  be  applied  to  DOE.  OA  being  placed  in  OD.  By 
reason  of  the  equality  of  the  circles  the  arc  AB  will  fall  in  DE. 

Conceive  the  angles  AOB  and  DOE  as  generated  by  a  radius  mov- 
ing from  the  position  -OA  (which  is  now  also  OD)  to  OB  and  OE, 
with  uniform  motion.     Let  the  time  of  generating  AOB  be  r,  and  that  of 

AOB      r 

generating  DOE  be  s.     Whence  —^  =  -  (48,  49). 

Again,  the  extremity  of  the  radius,  as  A  (or  D),  describes  equal  and  (as 
far  as  the  less  extends)  coincident  parts  of  AB  and  DE  in  equal  times,  whence 

arcAB  _   r     „  , ,...  _.  __.: ,,„_  AOB  _  arc  AB 

arc  DE 


Hence,  by  equality  of  ratios,  we  have  pQ£  =  a7c  qE 


2.  To  prove  the  same  proposition  by  the  infinitesimal  method,  we  pro 
ceed  exactly  as  in  Case  II.,  pp.  89,  90.  simply  conceiving  vi  as  infinitesimal 
when  the  angles  are  incommensurable  by  a  finite  unit,  and  for  5  putting  the 
indefinite  number  r,  and  for  8  the  indefinite  number  s. 


INFINITESIMAL    METHOD.  333 


NOTE    ON     (343). 

By  the  old  method  the  Lemma  on  which  this  demonstration  is  based  is 
proved  in  two  cases.  1st.  When  the  bases  are  commensurable  ;  2nd.  When 
the  bases  are  incommensurable.  Dividing  the  bases  into  equal  parts  and 
erecting  perpendiculars  at  the  points  of  division  the  argument  in  the  first 
case  proceeds  exactly  like  the  argument  in  Case  II.  of  (182).  When  the 
bases  are  incommensurable,  we  apply  abed  to  ABCD  placin«ir  ad  in  its 
equal  AD,  whence  ab  falls  in  AB,  as  far  as  it  extends,  and  dc  in  DC.     Then 

^,    ^   .-  ABCD  .        ^  ,       AB   .^  .  ,       AB         ,    .         .^, 

assume  that,  if  — ;; — ~  is  not  equal  to  -^r ,  it  is  equal  to  —  ,  ag  being  either 
abed  ^  ab  ag     ^  ° 

greater  or  less  than  ab.     Now  divide  the  base  AB  into  equal  parts,  each  of 

which  is   less  than   bg,  and  erect  perpendiculars  at  each  of  the  points  of 

division.     We  may  then  show,  as  in  Case  III.  of  (182),  the  absurdity  of 

supposing  ag  greater  or  less  than  ab. 


FINIS. 


'i^-A^ 


'^CX. 


'7- 


.'/^ 


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